Problem 15

Question

Find $y^{\prime}$. (a) $y=\frac{x}{\sin x}$ (b) $y=3 \tan ^{3}\left(x^{2}\right)$ (c) $y=\tan \left(\frac{x}{3}\right) \sec (3 x)$

Step-by-Step Solution

Verified

Answer

(a) $y' = \frac{\sin x - x \cos x}{\sin^2 x}$ (b) $y' = 18x \tan^2(x^2) \sec^2(x^2)$ (c) $y' = \frac{1}{3} \sec(x/3) \sec(3x) + 3 \tan(x/3) \sec(3x) \tan(3x)$

1Step 1: Differentiate $y=\frac{x}{\sin x}$

This requires the quotient rule. Here, $f = x$ (whose derivative, $f' = 1$) and $g = \sin x$ (whose derivative, $g' = \cos x)$. Applying the quotient rule yields: $y' = \frac{f'g - g'f}{g^2} = \frac{1*\sin x - \cos x * x}{(\sin x)^2} = \frac{\sin x - x \cos x}{\sin^2 x}$

2Step 2: Differentiate $y=3 \tan ^{3}\left(x^{2}\right)$

This requires the chain rule twice and remembering the derivative of $\tan x$ is $\sec^2 x$. Differentiate the outer function while keeping the inner function, $3 * 3* tan^{2}(x^2) = 9\tan^2(x^2)$, then multiply the result by the derivative of inner function (chain rule), $x^2$' = 2x. Finally, apply the derivative of $\tan x$, resulting in $y' = 2x * 9\tan^2(x^2) * \sec^2(x^2) = 18x \tan^2(x^2) \sec^2(x^2)$

3Step 3: Differentiate $y=\tan \left(\frac{x}{3}\right) \sec (3 x)$

This requires the product rule and chain rule. Differentiating $f = \tan(x/3)$ (where $f' = \frac{1}{3} \sec^2 (x/3)$, by the chain rule) and $g = \sec(3x)$ (where $g' = 3 \sec(3x) \tan(3x)$, by the chain rule), we apply the product rule to get: $y' = f'g + fg' = \frac{1}{3} \sec^2 (x/3)* \sec(3x) + \tan(x/3) * 3 \sec(3x) \tan(3x) = \frac{1}{3} \sec(x/3) \sec(3x) + 3 \tan(x/3) \sec(3x) \tan(3x)$

Key Concepts

Quotient RuleChain RuleProduct Rule

Quotient Rule

The quotient rule is a technique for finding the derivative of a division of two functions. When you have a function like $\frac{f(x)}{g(x)}$, and you need to take its derivative, the quotient rule comes in handy. This rule can be remembered by the formula:

$y' = \frac{f'g - g'f}{g^2}$

This formula states that the derivative of a quotient is given by the bottom function $g$ times the derivative of the top function $f'$, minus the top function $f$ times the derivative of the bottom function $g'$, all over the bottom function squared $g^2$.

In step 1 of the exercise, to differentiate $y = \frac{x}{\sin x}$, we take:

$f(x) = x$ with $f'(x) = 1$
$g(x) = \sin x$ with $g'(x) = \cos x$
Thus, $y' = \frac{1 \cdot \sin x - x \cdot \cos x}{(\sin x)^2}$

The quotient rule provides a systematic way to solve for derivatives of expressions where functions are divided.

Chain Rule

The chain rule is essential when differentiating composite functions, where one function sits inside another. To apply the chain rule, you take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function.

Consider in step 2 the differentiation of $y = 3 \tan^3(x^2)$. It involves chaining derivatives: first handle the cubic power, then the $\tan x$, and finally the $x^2$.

Differentiate the outer function: from $y = u^3$ to $3u^2$, substitute $u = \tan(x^2)$ getting $9 \tan^2(x^2)$.
Then multiply by the derivative of $\tan(x^2)$, using it as $\sec^2(x^2)$.
Finally, the derivative of the inner $x^2$ is $2x$.
Combine these to get $18x \tan^2(x^2) \sec^2(x^2)$.

The chain rule stitches together these derivatives along the downside path to calculate a composite's derivative.

Product Rule

The product rule is your tool for differentiating the product of two or more functions. For two functions $f(x)$ and $g(x)$, the derivative $y'$ of their product $f(x)g(x)$ is:

$y' = f'g + fg'$

It requires you to differentiate each function separately and then add the products of these derivatives to the original functions.

In step 3, you find the derivative of $y=\tan(\frac{x}{3})\sec(3x)$. First, differentiate $f = \tan(\frac{x}{3})$: