Question

Fx=xsinx
Derivative by first princpal​

Answer 1

Answer:

sinx + xcosx

Step-by-step explanation:

$\sf{f'(x) = \lim_{h \to 0} \dfrac{(x + h)sin(x + h) - xsinx}{h}}\\\\\sf{f'(x)= \lim_{h \to 0}\dfrac{(x+h)(sinx.cosh + sinh.cosx) - xsinx}{h}}\\\\\\\sf{f'(x) = \lim_{h \to 0} \dfrac{xsinx.cosh + xsinh.cosx + hsinx.cosh+hsinh.cosx - xsinx}{h} }\\\\\sf{f'(x)= \lim_{h \to 0}\bigg(\dfrac{xsinxcosh}{h} + x\dfrac{sinh}{h} cosx + sinxcosh+sinhcosx - \dfrac{xsinx}{h}\bigg)}$

$\sf{f'(x)= \lim_{h\to 0} \dfrac{xsinx(cosh-1)}{h} +\lim_{h\to 0}(x\dfrac{sinh}{h} cosx) + \lim_{h\to 0}sinxcosh+\lim_{h\to 0}(sinhcosx) }\\\\\sf{f'(x) = 0 + x(1)cosx + sinxcos(0) + sin(0)cosx}\\\\\sf{f'(x) = xcosx + sinx(1) + 0}\\\\\sf{f'(x) = sinx + xcosx}$

Answer 2

$\mathbb{ \ \ A\ N\ S\ W\ E\ R\: \: }$

• if y=f(x) g (x)
• then differentiation w.r.t.x
• dy/dx=f(x) dg(x)/dx+g(x)df(x)/dx
• Now Use This
• y=x.sinx
• dy/dx=x.d (sinx)/dx+sinx.dx/dx
• =x.cosx+sinx