Question

f(x)=x²cosx then f'(x)=?
using the formula,f'(x)=f(x+h)-f(x)/h
please use this☝️☝️​

Answer 1

Step-by-step explanation:

We have,

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

Now,

$\frac{d(f(x))}{dx} = \lim_{h \rarr0} \frac{f(x + h) - f(x)}{h}$

$= \lim_{h \rarr0} \frac{(x + h) ^{2} \cos(x + h) - {x}^{2} \cos(x) }{h}$

$= \lim_{h \rarr0} \frac{ {x}^{2}( \cos(x + h) - \cos(x) ) + h(2x + h) \cos(x + h) }{h}$

$= \lim_{h \rarr0} \frac{ {x}^{2}( - 2 \sin(x + \frac{h}{2} ) \sin( \frac{h}{2} ) )}{h} + \lim_{h \rarr0}(2x + h) \cos(x + h)$

$= - \lim_{h \rarr0} \sin(x + \frac{h}{2} ) \times \lim_{h \rarr0} \frac{ \sin( \frac{h}{2} ) }{ \frac{h}{2} } + (2x + 0) \cos(x + 0)$

$= - \sin(x) \times 1 + 2x \cos(x)$

$= 2x \cos(x) - \sin(x)$