Question Find the derivative of $f(x) = (x^{3} + x^{2})^{\tan x}$ using logarithmic differentiation.

Studdy Solution

STEP 1

Assumptions
1. The function to differentiate is $f(x) = \left(x^{3} + x^{2}\right)^{\tan x}$ .
2. We will use logarithmic differentiation which involves taking the natural logarithm of both sides of the equation $f(x)$ and then differentiating implicitly.
3. The rules of differentiation that will be used include the chain rule, product rule, and the derivative of natural logarithm.

STEP 2

Begin by taking the natural logarithm of both sides of the equation to simplify the differentiation process. Let $y = f(x)$ , then take the logarithm of $y$ .
$\ln(y) = \ln\left(\left(x^{3} + x^{2}\right)^{\tan x}\right)$

STEP 3

Apply the property of logarithms that allows us to bring the exponent in front as a multiplier.
$\ln(y) = \tan x \cdot \ln\left(x^{3} + x^{2}\right)$

STEP 4

Now, differentiate both sides of the equation with respect to $x$ . Remember to use the product rule on the right-hand side, where the product rule states that $(uv)' = u'v + uv'$ .
$\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[\tan x \cdot \ln\left(x^{3} + x^{2}\right)]$

STEP 5

Differentiate the left-hand side using the chain rule, where $\frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx}$ .
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[\tan x \cdot \ln\left(x^{3} + x^{2}\right)]$

STEP 6

Apply the product rule to the right-hand side.
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[\tan x] \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{d}{dx}[\ln\left(x^{3} + x^{2}\right)]$

STEP 7

Differentiate $\tan x$ and $\ln\left(x^{3} + x^{2}\right)$ separately.
$\frac{1}{y} \cdot \frac{dy}{dx} = \sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{1}{x^{3} + x^{2}} \cdot \frac{d}{dx}\left(x^{3} + x^{2}\right)$

STEP 8

Differentiate $x^{3} + x^{2}$ using the power rule, where $\frac{d}{dx}[x^n] = nx^{n-1}$ .
$\frac{1}{y} \cdot \frac{dy}{dx} = \sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{1}{x^{3} + x^{2}} \cdot \left(3x^{2} + 2x\right)$

STEP 9

Simplify the expression by combining like terms.
$\frac{1}{y} \cdot \frac{dy}{dx} = \sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{3x^{2} + 2x}{x^{3} + x^{2}}$

STEP 10

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$ , which is $f(x) = \left(x^{3} + x^{2}\right)^{\tan x}$ .
$\frac{dy}{dx} = y \cdot \left(\sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{3x^{2} + 2x}{x^{3} + x^{2}}\right)$

STEP 11

Substitute $y$ back with $f(x)$ .
$\frac{dy}{dx} = \left(x^{3} + x^{2}\right)^{\tan x} \cdot \left(\sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{3x^{2} + 2x}{x^{3} + x^{2}}\right)$

STEP 12

This is the final expression for the derivative of $f(x)$ .
$f'(x) = \left(x^{3} + x^{2}\right)^{\tan x} \cdot \left(\sec^{2}x \cdot \ln\left(x^{3} + x^{2}\right) + \tan x \cdot \frac{3x^{2} + 2x}{x^{3} + x^{2}}\right)$

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