If $y = \frac{u}{v}$, where $u$ and $v$ are differentiable functions of $x$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Derivation

Let $y = \frac{u(x)}{v(x)}$. If $\delta y$, $\delta u$, and $\delta v$ represent small increments in $y$, $u$, and $v$ respectively, then:

$y + \delta y = \frac{u + \delta u}{v + \delta v}$

$\delta y = \frac{u + \delta u}{v + \delta v} - \frac{u}{v}$

$\delta y = \frac{v(u + \delta u) - u(v + \delta v)}{v(v + \delta v)}$

$\delta y = \frac{vu + v\delta u - uv - u\delta v}{v(v + \delta v)}$

$\delta y = \frac{v\delta u - u\delta v}{v(v + \delta v)}$

Dividing by $\delta x$:

$\frac{\delta y}{\delta x} = \frac{v\frac{\delta u}{\delta x} - u\frac{\delta v}{\delta x}}{v(v + \delta v)}$

Taking the limit as $\delta x \to 0$ (which also implies $\delta v \to 0$):

$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{v\frac{\delta u}{\delta x} - u\frac{\delta v}{\delta x}}{v(v + \delta v)} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Example 1

Find the derivative of $y = \frac{x^2 + 4}{x^2 - 4}$.

Solution:

Let $u = x^2 + 4$ and $v = x^2 - 4$.

$\frac{du}{dx} = 2x \quad \text{and} \quad \frac{dv}{dx} = 2x$

Using the quotient rule:

$\frac{dy}{dx} = \frac{(x^2 - 4)(2x) - (x^2 + 4)(2x)}{(x^2 - 4)^2}$

$= \frac{2x^3 - 8x - 2x^3 - 8x}{(x^2 - 4)^2}$

$= \frac{-16x}{(x^2 - 4)^2}$

Example 2

If $y = \frac{x^3 + 5x^2 + 2}{x + 1}$, find $\frac{dy}{dx}$.

Solution:

Let $u = x^3 + 5x^2 + 2$ and $v = x + 1$.

$\frac{du}{dx} = 3x^2 + 10x \quad \text{and} \quad \frac{dv}{dx} = 1$

Using the quotient rule:

$\frac{dy}{dx} = \frac{(x + 1)(3x^2 + 10x) - (x^3 + 5x^2 + 2)(1)}{(x + 1)^2}$

$= \frac{3x^3 + 10x^2 + 3x^2 + 10x - x^3 - 5x^2 - 2}{(x + 1)^2}$

$= \frac{2x^3 + 8x^2 + 10x - 2}{(x + 1)^2}$

The chain rule

The chain rule is used to differentiate composite functions (functions within functions). If $y$ is a function of $u$, say $y = f(u)$, and $u$ is a function of $x$, say $u = g(x)$, then $y$ is a composite function of $x$.

The chain rule states:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Derivation

Let $y = f(u)$ and $u = g(x)$. If $\delta y$ and $\delta u$ are small increments in $y$ and $u$ respectively, then:

$y + \delta y = f(u + \delta u)$

$\delta y = f(u + \delta u) - f(u)$

Dividing both sides by $\delta x$:

$\frac{\delta y}{\delta x} = \frac{f(u + \delta u) - f(u)}{\delta x}$

$\frac{\delta y}{\delta x} = \frac{f(u + \delta u) - f(u)}{\delta u} \cdot \frac{\delta u}{\delta x}$

Taking the limit as $\delta x \to 0$ (which implies $\delta u \to 0$):

$\lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta u \to 0} \frac{f(u + \delta u) - f(u)}{\delta u} \cdot \lim_{\delta x \to 0} \frac{\delta u}{\delta x}$

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Example 1

Given $y = (x^2 + 2)^6$, find $\frac{dy}{dx}$.

Solution:

Let $u = x^2 + 2$, then $y = u^6$.

$\frac{du}{dx} = 2x \quad \text{and} \quad \frac{dy}{du} = 6u^5$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6u^5 \cdot 2x = 12x u^5$

Substituting $u = x^2 + 2$ back in:

$\frac{dy}{dx} = 12x(x^2 + 2)^5$

Example 2

If $y = (x^3 - 3x + 2)^6$ and $u = x^3 - 3x + 2$, express $y$ in terms of $u$. Hence, find $\frac{dy}{dx}$ at $x = 2$.

Solution:

Given $y = (x^3 - 3x + 2)^6$ and $u = x^3 - 3x + 2$, then $y = u^6$.

$\frac{dy}{du} = 6u^5 \quad \text{and} \quad \frac{du}{dx} = 3x^2 - 3$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6u^5(3x^2 - 3)$

Substituting $u$ back in:

$\frac{dy}{dx} = 6(x^3 - 3x + 2)^5(3x^2 - 3)$

At $x = 2$:

$\frac{dy}{dx} = 6(2^3 - 3(2) + 2)^5(3(2)^2 - 3) = 6(8 - 6 + 2)^5(12 - 3) = 6(4)^5(9) = 6(1024)(9) = 55296$

Example 3

Find the derivative of $y = (4x^2 - 3x + 4)^{\frac{3}{2}}$ at $x = 1$.

Solution:

Let $u = 4x^2 - 3x + 4$, then $y = u^{\frac{3}{2}}$.

$\frac{du}{dx} = 8x - 3 \quad \text{and} \quad \frac{dy}{du} = \frac{3}{2}u^{\frac{1}{2}}$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{3}{2}u^{\frac{1}{2}}(8x - 3)$

$\frac{dy}{dx} = \frac{3}{2}(4x^2 - 3x + 4)^{\frac{1}{2}}(8x - 3)$

At $x = 1$:

$\frac{dy}{dx} = \frac{3}{2}(4(1)^2 - 3(1) + 4)^{\frac{1}{2}}(8(1) - 3) = \frac{3}{2}(5)^{\frac{1}{2}}(5) = \frac{15\sqrt{5}}{2} \approx 16.77$

Differentiation of implicit functions

An implicit function is a function where the dependent variable ($y$) is not explicitly expressed in terms of the independent variable ($x$). It has the form $f(x, y) = 0$. For example, $x^2 + y^2 = 25$ is an implicit function.

Implicit differentiation is a technique used to find $\frac{dy}{dx}$ when $y$ is defined implicitly as a function of $x$. The key is to differentiate both sides of the equation with respect to $x$, using the chain rule when differentiating terms involving $y$.

Remember that $\frac{d}{dx}f(y) = \frac{df}{dy} \cdot \frac{dy}{dx}$.

Also, the derivative of a term like $x^m y^n$ is given by:

$\frac{d}{dx}(x^m y^n) = mx^{m-1}y^n + nx^m y^{n-1} \frac{dy}{dx}$

Example 1

Differentiate $z = y^2$ with respect to $x$.

Solution:

Differentiating both sides with respect to $x$:

$\frac{dz}{dx} = \frac{d}{dx}(y^2)$

Using the chain rule:

$\frac{dz}{dx} = 2y \frac{dy}{dx}$

Example 2

Find $\frac{dy}{dx}$ for the curve $3x^2 - 4y^2 = 12$ at the point $(4, 3)$.

Solution:

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(3x^2) - \frac{d}{dx}(4y^2) = \frac{d}{dx}(12)$

$6x - 8y\frac{dy}{dx} = 0$

$8y\frac{dy}{dx} = 6x$

$\frac{dy}{dx} = \frac{6x}{8y} = \frac{3x}{4y}$

At the point $(4, 3)$:

$\frac{dy}{dx} = \frac{3(4)}{4(3)} = \frac{12}{12} = 1$

Example 3

If $y = \frac{x^2 - 6}{2x + 3}$, show that $\left(\frac{dy}{dx}\right)^2 = \frac{225}{(4x - 6)^3}$.

Solution:

First, find $\frac{dy}{dx}$ using the quotient rule:

$\frac{dy}{dx} = \frac{(2x + 3)(2x) - (x^2 - 6)(2)}{(2x + 3)^2}$

$\frac{dy}{dx} = \frac{4x^2 + 6x - 2x^2 + 12}{(2x + 3)^2}$

$\frac{dy}{dx} = \frac{2x^2 + 6x + 12}{(2x + 3)^2}$

This is where the provided original solution had an error. Let's recalculate the derivative correctly and then square it:

$\frac{dy}{dx} = \frac{2(x^2 + 3x + 6)}{(2x + 3)^2}$

Now, let's consider the given expression $y = \frac{x^2 - 6}{2x + 3}$. If we were to calculate $y^2$, we get:

$y^2 = \frac{(x^2 - 6)^2}{(2x+3)^2}$

It appears there's a misunderstanding in the original problem statement or the expected result. The provided "show that" statement does not directly follow from the given function using standard differentiation. The correct derivative is calculated above. The given solution seems to be attempting to relate the square of the derivative to an expression involving the original function in a way that isn't mathematically sound.

Therefore, the correct derivative is $\frac{dy}{dx} = \frac{2x^2 + 6x + 12}{(2x + 3)^2}$ or simplified $\frac{2(x^2 + 3x + 6)}{(2x + 3)^2}$. The "show that" part of the original problem is incorrect.

Further implicit differentiation

Implicit differentiation can be performed without explicitly solving for one variable in terms of the other. We differentiate each term with respect to $x$, applying the chain rule and product rule as needed.

Example 1

Find $\frac{dy}{dx}$ for $2y^3 + xy^2 - y + 5 = 4$.

Solution:

Differentiating each term with respect to $x$:

$\frac{d}{dx}(2y^3) + \frac{d}{dx}(xy^2) - \frac{d}{dx}(y) + \frac{d}{dx}(5) = \frac{d}{dx}(4)$

Applying the chain rule and product rule:

$6y^2\frac{dy}{dx} + (1 \cdot y^2 + x \cdot 2y\frac{dy}{dx}) - \frac{dy}{dx} + 0 = 0$

$6y^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} - \frac{dy}{dx} = 0$

Collecting terms with $\frac{dy}{dx}$:

$(6y^2 + 2xy - 1)\frac{dy}{dx} = -y^2$

$\frac{dy}{dx} = \frac{-y^2}{6y^2 + 2xy - 1}$

Example 2

If $5x^3 + 4xy^3 - 5y^2 = 15$, find $\frac{dy}{dx}$.

Solution:

Differentiating each term with respect to $x$:

$\frac{d}{dx}(5x^3) + \frac{d}{dx}(4xy^3) - \frac{d}{dx}(5y^2) = \frac{d}{dx}(15)$

Applying the product and chain rules:

$15x^2 + (4y^3 + 4x \cdot 3y^2\frac{dy}{dx}) - 10y\frac{dy}{dx} = 0$

$15x^2 + 4y^3 + 12xy^2\frac{dy}{dx} - 10y\frac{dy}{dx} = 0$

Collecting terms with $\frac{dy}{dx}$:

$(12xy^2 - 10y)\frac{dy}{dx} = -15x^2 - 4y^3$

$\frac{dy}{dx} = \frac{-15x^2 - 4y^3}{12xy^2 - 10y} = \frac{15x^2 + 4y^3}{10y-12xy^2}$

Example 3

Find the gradient of the curve $x^2 - 2xy + y^2 = 1$ at the point $(2, 1)$.

Solution:

Differentiating each term with respect to $x$:

$\frac{d}{dx}(x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1)$

$2x - (2y + 2x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$

$2x - 2y - 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

Collecting terms with $\frac{dy}{dx}$:

$(2y - 2x)\frac{dy}{dx} = 2y - 2x$

$\frac{dy}{dx} = \frac{2y - 2x}{2y - 2x} = 1, \quad \text{provided } y \ne x$

At the point $(2, 1)$:

$\frac{dy}{dx} = \frac{2(1) - 2(2)}{2(1) - 2(2)} = \frac{-2}{-2} = 1$

Derivatives of trigonometric functions

This section covers the differentiation of trigonometric functions, primarily using first principles and the chain rule.

Example 1: Derivative of $f(x) = \sin x$

Using the first principles definition:

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

Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:

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

$f'(x) = \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$

$f'(x) = \sin x \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \lim_{h \to 0} \frac{\sin h}{h}$

Since $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ and $\lim_{h \to 0} \frac{\sin h}{h} = 1$:

$f'(x) = \sin x (0) + \cos x (1) = \cos x$

Therefore, $\frac{d}{dx}(\sin x) = \cos x$.

Example 2: Derivative of $y = \cos x$

Using the first principles definition:

$\frac{dy}{dx} = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h}$

Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:

$\frac{dy}{dx} = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$

$\frac{dy}{dx} = \lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h}$

$\frac{dy}{dx} = \cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \lim_{h \to 0} \frac{\sin h}{h}$

Using the same limits as above:

$\frac{dy}{dx} = \cos x (0) - \sin x (1) = -\sin x$

Therefore, $\frac{d}{dx}(\cos x) = -\sin x$.

Example 3: Derivative of $\sin 3\theta$

Let $y = \sin 3\theta$. Let $u = 3\theta$, so $y = \sin u$.

$\frac{du}{d\theta} = 3 \quad \text{and} \quad \frac{dy}{du} = \cos u$

Using the chain rule:

$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} = \cos u \cdot 3 = 3\cos 3\theta$

Example 4: Gradient of $y = 5\sin x - x^2$ at $x = -\pi$

$\frac{dy}{dx} = 5\cos x - 2x$

At $x = -\pi$:

$\frac{dy}{dx} = 5\cos(-\pi) - 2(-\pi) = 5(-1) + 2\pi = 2\pi - 5$

Example 5: Derivative of $\sec \theta$ from first principles

Let $f(\theta) = \sec \theta = \frac{1}{\cos \theta}$.

$f'(\theta) = \lim_{h \to 0} \frac{\frac{1}{\cos(\theta + h)} - \frac{1}{\cos \theta}}{h}$

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

Using the trigonometric identity $\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$:

$f'(\theta) = \lim_{h \to 0} \frac{2\sin(\frac{2\theta + h}{2})\sin(\frac{h}{2})}{h \cos \theta \cos(\theta + h)}$

$f'(\theta) = \lim_{h \to 0} \frac{\sin(\theta + \frac{h}{2})}{\cos \theta \cos(\theta + h)} \cdot \frac{\sin(\frac{h}{2})}{\frac{h}{2}}$

As $h \to 0$, $\frac{\sin(\frac{h}{2})}{\frac{h}{2}} \to 1$, so:

$f'(\theta) = \frac{\sin \theta}{\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta} = \tan \theta \sec \theta$

Example 6: Derivative of $y = \cos(\frac{1}{x})$

Let $u = \frac{1}{x} = x^{-1}$, so $y = \cos u$.

$\frac{du}{dx} = -x^{-2} = -\frac{1}{x^2} \quad \text{and} \quad \frac{dy}{du} = -\sin u$

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (-\sin u)(-\frac{1}{x^2}) = \frac{1}{x^2}\sin(\frac{1}{x})$

Example 7: If $f(x) = \frac{\cos x}{x}$, show that $f'(x) = \frac{-x\sin x - \cos x}{x^2}$

Using the quotient rule:

$f'(x) = \frac{x(-\sin x) - (\cos x)(1)}{x^2} = \frac{-x\sin x - \cos x}{x^2}$

Derivatives of inverse trigonometric functions

This section covers the differentiation of inverse trigonometric functions.

1. Derivative of $y = \sin^{-1} x$

If $y = \sin^{-1} x$, then $x = \sin y$.

Differentiating both sides with respect to $x$:

$1 = \cos y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\cos y}$

Since $\cos^2 y = 1 - \sin^2 y$ and $\sin y = x$, we have $\cos y = \sqrt{1 - x^2}$.

Therefore, $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$.

Example 1: Differentiate $y = \sin^{-1}(2x - 3)$

Let $u = 2x - 3$, so $y = \sin^{-1} u$.

$\frac{du}{dx} = 2 \quad \text{and} \quad \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot 2 = \frac{2}{\sqrt{1 - (2x - 3)^2}} = \frac{2}{\sqrt{1 - (4x^2 - 12x + 9)}} = \frac{2}{\sqrt{-4x^2 + 12x - 8}}$

2. Derivative of $y = \cos^{-1} x$

If $y = \cos^{-1} x$, then $x = \cos y$.

Differentiating both sides with respect to $x$:

$1 = -\sin y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{-1}{\sin y}$

Since $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$:

$\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1 - x^2}}$

Example 2: Differentiate $y = \cos^{-1}(3x-1)$

Let $u = 3x - 1$, so $y = \cos^{-1} u$.

$\frac{du}{dx} = 3 \quad \text{and} \quad \frac{dy}{du} = \frac{-1}{\sqrt{1 - u^2}}$

Using the chain rule:

$\frac{dy}{dx} = \frac{-3}{\sqrt{1 - (3x - 1)^2}} = \frac{-3}{\sqrt{1 - (9x^2 - 6x + 1)}} = \frac{-3}{\sqrt{-9x^2 + 6x}}$

3. Derivative of $y = \tan^{-1} x$

If $y = \tan^{-1} x$, then $x = \tan y$.

Differentiating both sides with respect to $x$:

$1 = \sec^2 y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$

Since $\sec^2 y = 1 + \tan^2 y$ and $\tan y = x$:

$\frac{dy}{dx} = \frac{1}{1 + x^2}$

Example 3: Differentiate $y = \tan^{-1}(\frac{1}{x})$

Let $u = \frac{1}{x}$, so $y = \tan^{-1} u$.

$\frac{du}{dx} = -\frac{1}{x^2} \quad \text{and} \quad \frac{dy}{du} = \frac{1}{1 + u^2}$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{1 + (\frac{1}{x})^2} \cdot (-\frac{1}{x^2}) = \frac{1}{1 + \frac{1}{x^2}} \cdot (-\frac{1}{x^2}) = \frac{x^2}{x^2 + 1} \cdot (-\frac{1}{x^2}) = \frac{-1}{x^2 + 1}$

Example 4: If $y = \tan^{-1}(\frac{\cos x}{1 - \sin x})$, show that $\frac{dy}{dx} = \frac{1}{2}$

This example was quite complex in the original and contained errors. A more straightforward approach uses trigonometric identities to simplify first:

$\frac{\cos x}{1 - \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{1 - \cos(\frac{\pi}{2} - x)} = \frac{2\sin(\frac{\pi}{4} - \frac{x}{2})\cos(\frac{\pi}{4} - \frac{x}{2})}{2\sin^2(\frac{\pi}{4} - \frac{x}{2})} = \cot(\frac{\pi}{4} - \frac{x}{2}) = \tan(\frac{\pi}{4} + \frac{x}{2})$

Therefore $y = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$

So, $\frac{dy}{dx} = \frac{1}{2}$

Derivatives of logarithmic functions

Logarithmic functions are of the form $y = \log_b x$, where $x > 0$, $b > 0$, and $b \ne 1$. They are the inverse of exponential functions.

If $y = \ln x$ (which means $y = \log_e x$), then $e^y = x$.

Differentiating both sides with respect to $x$:

$e^y \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$

Therefore, $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

More generally, if $y = \ln(f(x))$, then $\frac{dy}{dx} = \frac{f'(x)}{f(x)}$.

Example 1: If $y = 3\log_e(x - 2)$, show that $\frac{dy}{dx} = \frac{3}{x - 2}$

Let $u = x - 2$, so $y = 3\log_e u = 3\ln u$.

$\frac{du}{dx} = 1 \quad \text{and} \quad \frac{dy}{du} = \frac{3}{u}$

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{3}{u} \cdot 1 = \frac{3}{x - 2}$

Example 2: Show that $\frac{d}{dx}(x\ln(\sin x)) = \ln(\sin x) + x\cot x$

Let $y = x\ln(\sin x)$.

Using the product rule:

$\frac{dy}{dx} = (1)\ln(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \ln(\sin x) + x\cot x$

Example 3: Given $y = \log_e(\frac{x-1}{x+1})$, evaluate $y'$ at $x = 2$.

$y = \ln(\frac{x-1}{x+1}) = \ln(x-1) - \ln(x+1)$

$y' = \frac{1}{x-1} - \frac{1}{x+1} = \frac{(x+1)-(x-1)}{(x-1)(x+1)} = \frac{2}{x^2 - 1}$

At $x=2$, $y' = \frac{2}{2^2 - 1} = \frac{2}{3}$

Example 4: Find $\frac{dy}{dx}$ if $ye^x = \sin x$

Taking natural logarithms of both sides: $\ln(ye^x) = \ln(\sin x)$

$\ln y + x = \ln(\sin x)$

Differentiating implicitly with respect to x:

$\frac{1}{y}\frac{dy}{dx} + 1 = \frac{\cos x}{\sin x}$

$\frac{dy}{dx} = y(\cot x - 1) = \frac{\sin x}{e^x}(\cot x - 1) = \sin x (\cot x - 1)e^{-x}$

Example 5: Prove that the derivative of $y = x^4 \ln(\sin x)$ is $y' = x^3(4\ln(\sin x) + x\cot x)$

Using the product rule:

$y' = 4x^3 \ln(\sin x) + x^4 \cdot \frac{1}{\sin x} \cdot \cos x = x^3(4\ln(\sin x) + x\cot x)$

Example 6: Find $y'$ if $y = \log_a x^2$

Using the change of base formula: $\log_a x^2 = \frac{\ln x^2}{\ln a} = \frac{2\ln x}{\ln a}$

Differentiating with respect to $x$:

$y' = \frac{2}{\ln a} \cdot \frac{1}{x} = \frac{2}{x\ln a}$

Derivatives of exponential functions

This section covers the differentiation of exponential functions.