Question

Use logarithmic differentiation to find the derivative of the function. y = x2cos x Part 1 of 4 Using properties of logarithms, we can rewrite ln y = ln(x2cos x) as

Answer 1

ANSWER

${y}^{'} = 2x \cos(x) - {x}^{2} \sin(x)$

EXPLANATION

The given function is

$y = {x}^{2} \cos(x)$

We take natural log of both sides;

$ln(y) = ln({x}^{2} \cos(x) )$

Recall and use the product rule of logarithms.

$ln(AB) = ln(A ) + ln(B)$

This implies that:

$ln(y) = ln({x}^{2} ) + ln( \cos(x) )$

$ln(y) = 2 ln({x} ) + ln( \cos(x) )$

We now differentiate implicitly to obtain;

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

Multiply through by y,

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

Substitute y=x²cosx to obtain;

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

Expand:

${y}^{'} = 2x \cos(x) - {x}^{2} \sin(x)$