ANSWER
[tex]{y}^{
'} = 2x \cos(x) - {x}^{2} \sin(x) [/tex]
EXPLANATION
The given function is
[tex]y = {x}^{2} \cos(x) [/tex]
We take natural log of both sides;
[tex] ln(y) = ln({x}^{2} \cos(x) )[/tex]
Recall and use the product rule of logarithms.
[tex] ln(AB) = ln(A ) + ln(B) [/tex]
This implies that:
[tex]ln(y) = ln({x}^{2} ) + ln( \cos(x) )[/tex]
[tex]ln(y) = 2 ln({x} ) + ln( \cos(x) )[/tex]
We now differentiate implicitly to obtain;
[tex] \frac{ {y}^{
'} }{y} = \frac{2}{x} - \frac{ \sin(x) }{ \cos(x) } [/tex]
Multiply through by y,
[tex] {y}^{
'} = y( \frac{2}{x} - \frac{ \sin(x) }{ \cos(x) ) })[/tex]
Substitute y=x²cosx to obtain;
[tex] {y}^{
'} = {x}^{2} \cos(x) ( \frac{2}{x} - \frac{ \sin(x) }{ \cos(x) ) } )[/tex]
Expand:
[tex]{y}^{
'} = 2x \cos(x) - {x}^{2} \sin(x) [/tex]
