Question

If $y=(\cos x)^{2x}$, then $\frac{dy}{dx}$ is equal to

• A. $2(\cos x)^{2x}(\sin x-x \tan x)$
• B. $2(\cos x)^{2x}[\log(\cos x)+x \tan x]$
• C. $2(\sin x)^{2x}[\log(\cos x)-x \tan x]$
• D. $2(\sin x)^{2x}x \cot x$
• E. $2(\cos x)^{2x}[\log(\cos x)-x \tan x]$

Hint:

Derivative Tip: When differentiating $x^x$ or $f(x)^{g(x)}$, never try to use the standard power rule. The power rule only works when the exponent is a constant integer or fraction!

Answer 1

Answer: (E)

Concept:
When dealing with a function where both the base and the exponent contain the variable $x$ (like $y = [u(x)]^{v(x)}$), we must use logarithmic differentiation. We take the natural log of both sides to bring the exponent down before applying the Product Rule.

Step 1: Take the natural logarithm of both sides.
Given $y = (\cos x)^{2x}$: $$\log y = \log\left((\cos x)^{2x}\right)$$ Using log properties, bring down the exponent: $$\log y = 2x \cdot \log(\cos x)$$

Step 2: Differentiate both sides with respect to x.
Apply implicit differentiation on the left, and the Product Rule on the right: $$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} [2x \cdot \log(\cos x)]$$

Step 3: Apply the Product Rule.
Product Rule: $(uv)^{\prime} = u^{\prime}v + uv^{\prime}$ Let $u = 2x$ and $v = \log(\cos x)$. $u^{\prime} = 2$ $v^{\prime} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$ Substitute back: $$\frac{1}{y} \frac{dy}{dx} = (2)[\log(\cos x)] + (2x)[-\tan x]$$

Step 4: Isolate $\frac{dy{dx}$.}
Multiply both sides by $y$ to solve for the derivative: $$\frac{dy}{dx} = y \cdot [2\log(\cos x) - 2x \tan x]$$

Step 5: Substitute y back and factor.
Replace $y$ with the original function $(\cos x)^{2x}$ and factor out the common 2: $$\frac{dy}{dx} = (\cos x)^{2x} \cdot 2[\log(\cos x) - x \tan x]$$ $$\frac{dy}{dx} = 2(\cos x)^{2x} [\log(\cos x) - x \tan x]$$ Hence the correct answer is (E) $2(\cos x)^{2x[\log(\cos x)-x \tan x]$}.