CONTINUITY AND DIFFERENTIABILITY

Chapter 5

Continuity and Differentiability

Big idea: a graph that can be drawn without lifting the pen is continuous. A graph that is also smooth, with no sharp corner, can be given a slope at every point. That slope is the derivative.

Think about the charge level of a phone battery during the day. It never jumps from 80% to 40% in one instant. It slides down through every value in between. We say the battery level is a continuous function of time.

Now think about the price of posting a parcel. A parcel of 499 g may cost one price, and a parcel of 501 g may cost a higher price, with nothing in between. The price jumps. We say the price is not continuous at 500 g.

A quick test for a graph: can you draw it without lifting your pen? If yes, it is continuous. If you must lift the pen, there is a break.

Continuity is about no breaks. Differentiability is about no sharp corners. A function that is differentiable at a point has a clear tangent line there, and the slope of that tangent is the derivative.

The pen test is good for pictures, but we also need a test that works from a formula. A function $f$ is continuous at a point $x = c$ when three things are true:

1. $f(c)$ exists (the function has a value at $c$);
2. the limit $\lim_{x \to c} f(x)$ exists. This means the left-hand limit (coming from smaller $x$) and the right-hand limit (coming from larger $x$) are equal;
3. that limit is equal to $f(c)$.

In one line: left-hand limit = right-hand limit = $f(c)$. If any one of the three fails, the function is discontinuous at $c$.

Is $f(x) = 3 x^{2} - 2 x + 1$ continuous at $x = 2$?

Solution. The value is $f(2) = 3(2)^2 - 2(2) + 1 = 9$. When $x$ approaches 2 from the left, $f(x)$ approaches 9. When $x$ approaches 2 from the right, $f(x)$ also approaches 9. So left limit = right limit = $f(2) = 9$, and $f$ is continuous at 2. (The same works at every point: a polynomial is continuous everywhere.)

Let $g(x) = \frac{x^{2} - 9}{x - 3}$ for $x \neq 3$. At $x = 3$ the formula gives $\frac{0}{0}$, so $g(3)$ does not exist. But for every $x \neq 3$ we can cancel: $x^2 - 9 = (x - 3)(x + 3)$, so $g(x) = x + 3$. As $x$ approaches 3, $g(x)$ approaches $6$.

The limit exists (it is 6), but $g(3)$ does not exist, so condition 1 fails and $g$ is discontinuous at 3. This is called a removable discontinuity: if we decide that $g(3) = 6$, the hole is filled and the function becomes continuous.

Define $h$ by $h(x) = \begin{cases} x + 2 & x < 1 \\ 2 x + 3 & x \ge 1 \end{cases}$ At $x = 1$: the value is $h(1) = 5$. From the left, $h(x) = x + 2$ approaches 3. From the right, $h(x) = 2x + 3$ approaches 5. The two limits are different (3 ≠ 5), so no single limit exists and $h$ is discontinuous at 1.

For which value of $a$ is $f(x) = \begin{cases} a x + 2 & x \le 2 \\ 3 x^{2} - 4 & x > 2 \end{cases}$ continuous at $x = 2$?

Solution. From the right, $3x^2 - 4$ approaches $3(2)^2 - 4 = 8$. From the left (and at $x = 2$), $ax + 2$ equals $2a + 2$. For continuity these must be equal: $2a + 2 = 8$, so $2a = 6$ and $a = 3$. Check: with $a = 3$ the left limit is $3(2) + 2 = 8$, the right limit is 8 and $f(2) = 8$.

A function is continuous on an interval if it is continuous at every point of the interval. We usually do not check each point. We use these known facts:

• Polynomials are continuous everywhere.
• Rational functions (a polynomial divided by a polynomial) are continuous everywhere except where the denominator is 0.
• $|x|$, $\sin x$, $\cos x$ and $e^x$ are continuous everywhere. $\ln x$ is continuous for $x > 0$.
• $\tan x = \frac{\sin x}{\cos x}$ is continuous except where $\cos x = 0$.

Example 5. Is $|x|$ continuous at 0? The value is $|0| = 0$. From either side the limit is $0$: $\lim_{x \to 0^-} |x| = 0$ and $\lim_{x \to 0^+} |x| = 0$. All three agree, so $|x|$ is continuous at 0. It has a corner there, but no break.

Example 6. Let $f(x) = \frac{\sin x}{x}$ for $x \neq 0$. We can show that $\lim_{x \to 0} f(x) = 1$ (a famous limit). So to make $f$ continuous at 0 we must define $f(0) = 1$.

1. Find the left and right limits of $x^{2} + 1$ at $x = 3$. Is it continuous at 3?
2. What value should be given to $\frac{x^2 - 25}{x - 5}$ at $x = 5$ to make it continuous there?

Answers

1. Both limits are 10 and $f(3) = 10$. Yes, it is continuous.
2. The limit is 10, so define the value at 5 as 10.

Big functions are built from small ones. The next rules tell us that if the small parts are continuous, so is the big function.

If $f$ and $g$ are continuous at $c$, then these are also continuous at $c$:

• $f + g$ and $f - g$;
• $k \cdot f$ for any number $k$;
• $f \cdot g$;
• $\frac{f}{g}$, provided $g(c) \neq 0$;
• the composite $f(g(x))$, provided $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ ("a continuous function of a continuous function").

These rules are true because limits behave well: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on.

Is $f(x) = \left(x^{2} + 1\right) \left(2 x - 3\right)$ continuous at $x = 2$? Both $x^2 + 1$ and $2x - 3$ are polynomials, so both are continuous. Their product is continuous. Check by value: $f(2) = (4 + 1)(4 - 3) = 5$, and the limit as $x \to 2$ is also 5.

Is $g(x) = \frac{x + 4}{x - 1}$ continuous at $x = 3$? The denominator is $3 - 1 = 2$, which is not 0, so yes. Its value there is $\frac{7}{2} = 3.5$. At $x = 1$ the denominator is 0, so $g$ is not defined and not continuous at 1.

Let $f(x) = \frac{x + 1}{x^2 - 5x + 6}$. The top is a polynomial, so the only trouble is where the bottom is 0. Since $x^2 - 5x + 6 = (x - 2)(x - 3)$, this happens at $x = 2$ and $x = 3$ (check: the bottom equals 0 and 0 there). So $f$ is continuous everywhere except at $x = 2$ and $x = 3$.

Is $\sin(x^2 + 1)$ continuous for all $x$? The inside, $x^2 + 1$, is a polynomial, so it is continuous. The outside, $\sin$, is continuous. A continuous function of a continuous function is continuous, so yes, for every real $x$.

1. Where is $\frac{3}{x^2 - 16}$ discontinuous?
2. Is $x^2 \sin x$ continuous for all real $x$?

Answers

1. At $x = 4$ and $x = -4$ (the bottom is 0 there).
2. Yes. It is a product of two continuous functions, $x^2$ and $\sin x$.

1. Show that $f(x) = 2 x^{2} - 3 x + 7$ is continuous at $x = -1$.
2. The function $f(x) = \frac{x^{2} - 16}{x - 4}$ is defined for $x \neq 4$. What value should be given to $f(4)$ so that $f$ is continuous at 4?
3. Is $f(x) = \begin{cases} 3 x - 1 & x < 2 \\ 5 & x = 2 \\ x^{2} + 1 & x > 2 \end{cases}$ continuous at $x = 2$?
4. Is $f(x) = \begin{cases} x + 3 & x \le 1 \\ 5 x - 3 & x > 1 \end{cases}$ continuous at $x = 1$?
5. Find the value of $k$ for which $f(x) = \begin{cases} k x^{2} & x \le 2 \\ 3 & x > 2 \end{cases}$ is continuous at $x = 2$.
6. Find $a$ and $b$ so that $f(x) = \begin{cases} 4 & x \le 1 \\ a x + b & 1 < x < 3 \\ 10 & x \ge 3 \end{cases}$ is continuous everywhere.
7. Find all points where $f(x) = \frac{1}{x^2 - 9}$ is discontinuous.
8. Is $f(x) = |x - 2|$ continuous at $x = 2$?
9. Let $f(x) = \frac{x}{|x|}$ for $x \neq 0$. Find the left and right limits at 0. Can $f(0)$ be defined so that $f$ is continuous at 0?
10. Where is $f(x) = \frac{x + 2}{x^2 - 5x + 6}$ discontinuous?

Answers

1. $f(-1) = 12$, and the limit from both sides is 12. They agree, so $f$ is continuous at $-1$.
2. $f(4) = 8$ (cancel $x - 4$ to get $x + 4$, which tends to 8).
3. Left limit 5, right limit 5, $f(2) = 5$. All equal, so yes, it is continuous at 2.
4. Left limit 4, right limit 2. They are different, so $f$ is not continuous at 1 (a jump of 2).
5. $4k = 3$, so $k = \frac{3}{4}$.
6. Continuity at 1 gives $a + b = 4$. Continuity at 3 gives $3a + b = 10$. Subtracting, $2a = 6$, so $a = 3$ and $b = 1$.
7. At $x = 3$ and $x = -3$, where the bottom is 0.
8. Yes. Both one-sided limits are 0 and 0, and $f(2) = 0$.
9. The left limit is -1 and the right limit is 1. They differ, so no value of $f(0)$ can make $f$ continuous.
10. At $x = 2$ and $x = 3$, where $x^2 - 5x + 6 = (x - 2)(x - 3)$ is 0.

Take the curve $y = x^2$ and the point $(1, 1)$ on it. Choose a second point a small step $h$ to the right: $(1 + h, (1 + h)^2)$. The straight line through the two points is a secant. Its slope is $\frac{(1 + h)^2 - 1}{h}$. For $h = 1, 0.5, 0.25$ the slopes are 3, 2.5, 2.25. As $h$ shrinks, the second point slides toward the first, and the slopes settle down to 2. The line through $(1, 1)$ with slope 2 is the tangent line.

The number that the secant slopes settle down to is the derivative of $f$ at $c$. In symbols:

$$f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h}$$

If this limit exists, we say $f$ is differentiable at $c$. The limit from the right ($h > 0$) is the right-hand derivative; the limit from the left ($h < 0$) is the left-hand derivative. The derivative exists exactly when both are equal.

Example 1. For $f(x) = x^2$ at $c = 3$: $\frac{(3 + h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h$, which approaches 6 as $h \to 0$. So $f'(3) = 6$.

Example 2. For $f(x) = \frac{1}{x}$ at $c = 2$: $\frac{\frac{1}{2 + h} - \frac{1}{2}}{h}$ approaches $-0.25$, which is $-\frac{1}{4}$. So $f'(2) = -\frac{1}{4}$.

Look at $f(x) = |x|$ at 0. The step quotient is $\frac{|h| - 0}{h} = \frac{|h|}{h}$. For $h > 0$ it equals 1, so the right-hand derivative is 1. For $h < 0$ it equals $-1$, so the left-hand derivative is $-1$. They differ, so $|x|$ is not differentiable at 0.

Reason in plain words: a tangent line can only exist if the curve has no break there. The reverse is false: $|x|$ is continuous at 0 but not differentiable at 0.

Example 3 (smooth join). Let $f(x) = x^2$ for $x \le 1$ and $f(x) = 2x - 1$ for $x > 1$. Both pieces give 1 at $x = 1$, so $f$ is continuous. Left-hand derivative: 2. Right-hand derivative: 2. They agree, so $f$ is differentiable at 1 and $f'(1) = 2$.

Example 4 (corner). Let $g(x) = x^2$ for $x \le 1$ and $g(x) = 3x - 2$ for $x > 1$. Both pieces give 1 at $x = 1$, so $g$ is continuous. But the left-hand derivative is 2 and the right-hand derivative is 3. They differ, so $g$ is not differentiable at 1.

1. Use the definition to find $f'(1)$ for $f(x) = x^2 + 3x$.
2. Can a function be differentiable at a point where it is not continuous?

Answers

1. $\frac{(1+h)^2 + 3(1+h) - 4}{h} = 5 + h$, which approaches 5. So $f'(1) = 5$.
2. No. Differentiable at a point always means continuous there.

Here are the basic derivatives and rules you will use all through this chapter.

Rules for combining functions. If $u$ and $v$ are differentiable functions of $x$:

• Sum: $(u + v)' = u' + v'$
• Product: $(uv)' = u'v + uv'$
• Quotient: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, where $v \neq 0$
• Chain rule: if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$y = \left(3 x^{2} + 1\right)^{4}$. Outside: something to the power 4, derivative $4(\cdot)^3$. Inside: $3x^2 + 1$, derivative $6x$. So $\frac{dy}{dx} = 4(3x^2 + 1)^3 \cdot 6x = 24 x \left(3 x^{2} + 1\right)^{3}$

$y = \sin\left(5 x^{2}\right)$. Outside: $\sin$, derivative $\cos$. Inside: $5x^2$, derivative $10x$. So $\frac{dy}{dx} = 10 x \cos\left(5 x^{2}\right)$

$y = \sqrt{x^{2} + 4}$. Outside: square root, derivative $\frac{1}{2\sqrt{\cdot}}$. Inside: $x^2 + 4$, derivative $2x$. So $\frac{dy}{dx} = \frac{2x}{2\sqrt{x^2 + 4}} = \frac{x}{\sqrt{x^{2} + 4}}$

$y = \cos^{3}\left(2 x\right)$ means $(\cos 2x)^3$. Outside: cube, derivative $3(\cdot)^2$. Middle: $\cos$, derivative $-\sin$. Inside: $2x$, derivative 2. Multiply all three: $\frac{dy}{dx} = 3\cos^2 2x \cdot (-\sin 2x) \cdot 2 = -6 \cos^{2}\left(2 x\right) \sin\left(2 x\right)$

Product: $y = x^{2} \sin x$. Then $y' = (2x)\sin x + x^2 \cos x = 2 x \sin x + x^{2} \cos x$.

Quotient: $y = \frac{x}{x^{2} + 1}$. Then $y' = \frac{1 \cdot (x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \frac{1 - x^{2}}{\left(x^{2} + 1\right)^{2}}$.

1. Differentiate $\left(4 x - 5\right)^{3}$ with respect to $x$.
2. Differentiate $\cos\left(3 x + 1\right)$ with respect to $x$.
3. Differentiate $\sqrt{1 + x^{2}}$ with respect to $x$.
4. Differentiate $\sin^{3} x$ with respect to $x$.
5. Differentiate $\frac{2 x + 1}{x^{2} + 3}$ with respect to $x$.
6. Differentiate $x^{3} \cos x$ with respect to $x$.
7. Differentiate $\sin\left(\cos x\right)$ with respect to $x$.
8. Differentiate $\tan\left(4 x\right)$ with respect to $x$.

Answers

1. $12 \left(4 x - 5\right)^{2}$
2. $-3 \sin\left(3 x + 1\right)$
3. $\frac{x}{\sqrt{1 + x^{2}}}$
4. $3 \sin^{2} x \cos x$
5. $\frac{2 \left(x^{2} + 3\right) - \left(2 x + 1\right) \cdot 2 x}{\left(x^{2} + 3\right)^{2}}$
6. $3 x^{2} \cos x - x^{3} \sin x$
7. $-\sin x \cos\left(\cos x\right)$
8. $\frac{4}{\cos^{2}\left(4 x\right)}$

Often $x$ and $y$ are tied together by an equation, such as $x^2 + y^2 = 25$, and $y$ is not written alone on one side. We say $y$ is an implicit function of $x$. We can still find $\frac{dy}{dx}$ without solving for $y$.

Method.

1. Differentiate both sides with respect to $x$. Treat $y$ as a function of $x$, so the derivative of $y^2$ is $2y \frac{dy}{dx}$ (chain rule), and the derivative of $xy$ is $y + x\frac{dy}{dx}$ (product rule).
2. Collect all terms that contain $\frac{dy}{dx}$ on one side.
3. Solve for $\frac{dy}{dx}$.

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$. Differentiate: $2x + 2y\frac{dy}{dx} = 0$. So $\frac{dy}{dx} = \frac{-x}{y}$. At the point $(3, 4)$, which lies on the circle because $9 + 16 = 25$, the slope is $-\frac{3}{4}$ (that is -0.75).

Find $\frac{dy}{dx}$ if $x^2 + xy + y^2 = 7$. Differentiate each term: $2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$. Collect: $(x + 2y)\frac{dy}{dx} = -(2x + y)$. So $\frac{dy}{dx} = \frac{-\left(2 x + y\right)}{x + 2 y}$ The point $(1, 2)$ is on the curve ($1 + 2 + 4 = 7$) and the slope there is -0.8, which is $-\frac{4}{5}$.

For $x^3 + y^3 = 3xy$: $3x^2 + 3y^2 \frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$. Collect: $(y^2 - x)\frac{dy}{dx} = y - x^2$. So $\frac{dy}{dx} = \frac{y - x^{2}}{y^{2} - x}$ The point $(1.5, 1.5)$ is on the curve (both sides equal 6.75) and the slope there is -1.

1. Find $\frac{dy}{dx}$ for $x^2 + y^2 = 100$ at the point $(6, 8)$.

Answers

1. $\frac{dy}{dx} = -\frac{x}{y}$, which at $(6, 8)$ equals $-0.75$, that is $-\frac{3}{4}$. (Check: $36 + 64 = 100$.)

The function $\sin^{-1} x$ answers the question "which angle has sine equal to $x$?". It is defined for $-1 \le x \le 1$. Its derivative follows from implicit differentiation.

Why the formula is true. Let $y = \sin^{-1} x$. Then $x = \sin y$. Differentiate both sides with respect to $x$: $1 = \cos y \cdot \frac{dy}{dx}$. So $\frac{dy}{dx} = \frac{1}{\cos y}$. For the principal value, $\cos y \ge 0$, so $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. The other two formulas follow in the same way.

$y = \tan^{-1}(2x)$. Outside: $\tan^{-1}$, derivative $\frac{1}{1 + (\cdot)^2}$. Inside: $2x$, derivative 2. So $\frac{dy}{dx} = \frac{2}{1 + 4x^2}$, that is $\frac{2}{1 + 4 x^{2}}$.

$y = \sin^{-1}(x^2)$. The derivative is $\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x = \frac{2 x}{\sqrt{1 - x^{4}}}$, valid for $-1 < x < 1$.

Let $y = \tan^{-1}\left(\frac{1 + x}{1 - x}\right)$ for $x \neq 1$. The chain rule gives $\frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 + x}{1 - x}\right)^2} \cdot \frac{2}{(1 - x)^2}$. Simplify: the bottom of the first fraction is $\frac{(1 - x)^2 + (1 + x)^2}{(1 - x)^2} = \frac{2 + 2x^2}{(1 - x)^2}$. So the whole product is $\frac{2}{2 + 2x^2} = \frac{1}{1 + x^{2}}$. This is the same as the derivative of $\tan^{-1} x$. (That is because the two functions differ only by a constant, $\frac{\pi}{4}$, on each side of $x = 1$.)

1. Find $\frac{dy}{dx}$ if $x^2 + 3y^2 = 7$, and evaluate it at $(2, 1)$.
2. Find $\frac{dy}{dx}$ if $xy = 6$.
3. Find $\frac{dy}{dx}$ if $x^2 y + y^2 = x + 9$, and evaluate it at $(0, 3)$.
4. Differentiate $\sin^{-1}\left(3 x\right)$ with respect to $x$.
5. Differentiate $\tan^{-1}\left(x^{2}\right)$ with respect to $x$.
6. Differentiate $\cos^{-1}\left(\frac{x}{2}\right)$ with respect to $x$.
7. Differentiate $x \tan^{-1} x$ with respect to $x$.
8. Differentiate $\tan^{-1}\left(\frac{1 - x}{1 + x}\right)$ with respect to $x$.

Answers

1. $\frac{dy}{dx} = \frac{-x}{3 y}$. At $(2, 1)$ it equals $-\frac{2}{3}$ (-0.6667).
2. $\frac{dy}{dx} = \frac{-y}{x}$.
3. $\frac{dy}{dx} = \frac{-\left(2 x y - 1\right)}{x^{2} + 2 y}$. At $(0, 3)$ it equals $\frac{1}{6}$ (0.1667).
4. $\frac{3}{\sqrt{1 - 9 x^{2}}}$
5. $\frac{2 x}{1 + x^{4}}$
6. $\frac{-1}{\sqrt{4 - x^{2}}}$
7. $\tan^{-1} x + \frac{x}{1 + x^{2}}$
8. $\frac{-1}{1 + x^{2}}$

The number $e$ is about 2.71828. The function $e^x$ is special because its slope at every point equals its height at that point. The table below uses computed values:

In symbols: $\frac{d}{dx} e^x = e^x$. The natural logarithm $\ln x$ is the inverse function of $e^x$: it answers "what power of $e$ gives $x$?". It is defined only for $x > 0$.

Why $\frac{d}{dx}\ln x = \frac{1}{x}$: let $y = \ln x$. Then $x = e^y$. Differentiate both sides with respect to $x$: $1 = e^y \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$.

$y = e^{3 x^{2}}$. Outside: $e^{(\cdot)}$. Inside: $3x^2$, derivative $6x$. So $\frac{dy}{dx} = 6 x e^{3 x^{2}}$.

$y = \ln\left(x^{2} + 1\right)$. Outside: $\ln(\cdot)$, derivative $\frac{1}{\cdot}$. Inside: $x^2 + 1$, derivative $2x$. So $\frac{dy}{dx} = \frac{2 x}{x^{2} + 1}$.

$y = xe^x$. Product rule: $y' = 1 \cdot e^x + x \cdot e^x = \left(x + 1\right) e^{x}$.

$y = \ln(\sin x)$, for $x$ where $\sin x > 0$. Then $\frac{dy}{dx} = \frac{\cos x}{\sin x} = \frac{\cos x}{\sin x}$ (this is $\cot x$).

$y = 5^{x^2}$. Use $\frac{d}{dx}a^u = a^u \ln a \cdot u'$ with $a = 5$ and $u = x^2$. So $\frac{dy}{dx} = 2 x \ln 5 \cdot 5^{x^{2}}$.

1. Differentiate $e^{-x}$ with respect to $x$.
2. Differentiate $e^{x^{2} + 1}$ with respect to $x$.
3. Differentiate $\ln\left(3 x + 2\right)$ with respect to $x$.
4. Differentiate $x^{2} \ln x$ with respect to $x$.
5. Differentiate $\frac{e^{x}}{x}$ with respect to $x$.
6. Differentiate $3^{2 x}$ with respect to $x$.
7. Differentiate $\ln\left(\cos x\right)$ with respect to $x$.
8. Differentiate $e^{\sin x}$ with respect to $x$.

Answers

1. $-e^{-x}$
2. $2 x e^{x^{2} + 1}$
3. $\frac{3}{3 x + 2}$
4. $2 x \ln x + x$
5. $\frac{\left(x - 1\right) e^{x}}{x^{2}}$
6. $2 \ln 3 \cdot 3^{2 x}$
7. $\frac{-\sin x}{\cos x}$
8. $\cos x e^{\sin x}$

Some functions are hard to differentiate directly: a variable raised to a variable power, or a long product and quotient. The logarithm turns powers into multiplication and products into sums. This gives a method called logarithmic differentiation.

Method.

1. Take the natural logarithm of both sides, and use log laws to simplify: $\ln(ab) = \ln a + \ln b$, $\ln\frac{a}{b} = \ln a - \ln b$ and $\ln(a^b) = b\ln a$.
2. Differentiate both sides with respect to $x$. The left side $\ln y$ gives $\frac{1}{y}\frac{dy}{dx}$.
3. Multiply by $y$ and replace $y$ by its original expression.

Let $y = x^x$ for $x > 0$. Then $\ln y = x \ln x$. Differentiate: $\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$. So $\frac{dy}{dx} = x^{x} \left(\ln x + 1\right)$

Let $y = (\sin x)^x$ where $\sin x > 0$. Then $\ln y = x \ln(\sin x)$. Differentiate: $\frac{1}{y}\frac{dy}{dx} = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x}$. So $\frac{dy}{dx} = (\sin x)^x \left[\ln(\sin x) + x\cot x\right]$

Let $y = \frac{\left(x + 1\right)^{2} \left(x - 2\right)^{3}}{\left(x + 3\right)^{4}}$. Take logarithms: $\ln y = 2\ln(x + 1) + 3\ln(x - 2) - 4\ln(x + 3)$. Differentiate: $\frac{1}{y}\frac{dy}{dx} = \frac{2}{x + 1} + \frac{3}{x - 2} - \frac{4}{x + 3}$. So $\frac{dy}{dx} = y\left[\frac{2}{x + 1} + \frac{3}{x - 2} - \frac{4}{x + 3}\right]$ with $y$ replaced by the original product.

Let $y = x^{\sin x}$ for $x > 0$. Then $\ln y = \sin x \cdot \ln x$, so $\frac{1}{y}\frac{dy}{dx} = \cos x \ln x + \frac{\sin x}{x}$. Hence $\frac{dy}{dx} = x^{\sin x} \left(\cos x \ln x + \frac{\sin x}{x}\right)$