Question

Find the derivative of f(x) using the first principle of derivatives, where f(x) is xsinx.

Answer 1

Given:

f(x) = xsinx

To find:

Derivative of f(x)

First principle of derivative,

$\displaystyle{\sf{ f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x) }{h}}}$

Here,

$f(x) = xsinx \\ f(x+h) = (x+h)sin(x+h)$

Substituting,

$\displaystyle{\sf{ f'(x) = \lim_{h \to 0} \dfrac{ (x+h) sin (x+h) - x sin x }{h}}}$

Using Identity,

sin(A + B) = sinAcosB + cosAsinB

(A = x; B = h)

$\displaystyle{\sf{ f'(x) = \lim_{h \to 0} \dfrac{ (x+h)(sin \: x \: cos \: h+cos \: h\: sin \: x) - x sin x }{h}}}\\ \\ \displaystyle{\sf{ f'(x) = \lim_{h \to 0} \dfrac{ (xsin \: xcos\: h + cos\: xsin \: h) +h(sin \: x \: cos \: h+cos \: x \: sin \: h)-xsinx}{h}}} \\ \displaystyle{\sf{ f'(x) = \lim_{h \to 0} \dfrac{ (xsin \: x(cos\: h-1)+xcos\: xsin \: h+h(sin \: x \: cos \: h+sin \: h \: cos \: x) }{h}}}$

Cancel out the 'h' and seperate the limits,

$\displaystyle{\sf{f'(x) = \lim_{h \to 0} (\dfrac{ xsinx(cos \: h-1)}{h} + \dfrac{ xcos \: x sin \: h}{h} + \dfrac{ h(sinxcos \:h + cosx \:sin \:h)}{h})}}$

Removing constants 'xsin x' & 'xcos x' out and simplifying using rules of limits:

$\displaystyle{ \sf{f'(x) = -xsinx\lim_{h \to 0} \dfrac{(1 -cos\: h)}{h} + xcosx\lim_{h \to 0} \dfrac{sin \: h}{h} + \lim_{h \to 0}(sinxcos \:h + cosx \:sin \:h})} \\ \displaystyle{\sf{f'(x) = -xsinx(0)+xcosx(1)+(sincos0+cosxsin0)}} \\ \displaystyle{\sf{ f'(x) = xcosx+(sin(1)+cosx(0))}} \\ \displaystyle{\sf{f'(x) = xcos \: x+sin \: x}}$

Answer 2

Question=Find the derivative of f(x) using the first principle of derivatives, where f(x) is xsinx.

Solution⬇️

Given a function f(x)=x sin(x), and we have to find its derivative by the definition. Consider the expression (f(x+Delta x)-f(x))/(Delta x) and find its limit for Delta x->0:

(f(x+Delta x)-f(x))/(Delta x)=((x+Delta x)sin(x+Delta x)-x sin(x))/(Delta x)=

=(x(sin(x+Delta x)-sin(x))+Delta x sin(x+Delta x))/(Delta x)=

=x(sin(x+Delta x)-sin(x))/(Delta x)+ sin(x+Delta x).

The second summand has the limit sin(x), because sin(x) is continuous for any x.

The first summand has the limit x(sin(x))'=xcos(x) (by the definition of the derivative