Question

Differentiate using first principal
f(x) = x sinx​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\red{\rm :\longmapsto\:f(x) = x \: sinx}$

So,

$\red{\rm :\longmapsto\:f(x + h) = (x + h) \: sin(x + h)}$

Using Definition of First Principal, we have

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

On substituting the values, we get

$\sf \: = \: \displaystyle\lim_{h \to 0} \frac{(x + h)sin(x + h) - xsinx}{h}$

$\sf \: = \displaystyle\lim_{h \to 0} \frac{xsin(x + h) + hsin(x + h) - xsinx}{h}$

$\sf \: = \: \displaystyle\lim_{h \to 0} \frac{x\bigg[sin(x + h) - sinx\bigg]}{h} + \displaystyle\lim_{h \to 0} \frac{hsin(x + h)}{h}$

We know,

$\red{\boxed{ \rm{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, using this identity, we get

$\sf \: = \: x\displaystyle\lim_{h \to 0} \dfrac{2cos\bigg[\dfrac{x + h + x}{2} \bigg]sin\bigg[\dfrac{x + h - x}{2} \bigg]}{h} + sinx$

$\sf \: = \: x\displaystyle\lim_{h \to 0} \dfrac{2cos\bigg[\dfrac{2x + h}{2} \bigg]sin\bigg[\dfrac{h}{2} \bigg]}{h} + sinx$

$\sf \: = \:2 x\displaystyle\lim_{h \to 0} cos\bigg[\dfrac{2x + h}{2} \bigg] \times \displaystyle\lim_{h \to 0} \dfrac{sin\bigg[\dfrac{h}{2} \bigg]}{h} + sinx$

$\sf \: = \: 2x cos\bigg[\dfrac{2x + 0}{2} \bigg] \times \displaystyle\lim_{h \to 0} \dfrac{sin\bigg[\dfrac{h}{2} \bigg]}{\dfrac{h}{2} \times 2 } + sinx$

We know,

$\red{\boxed{ \rm{ \: \displaystyle\lim_{h \to 0} \frac{sinh}{h} = 1}}}$

So, using this, we get

$\sf \: = \: 2xcosx \times \dfrac{1}{2} + sinx$

$\sf \: = \: x \: cosx \: + \: sinx$

$\bf\implies \: \red{\boxed{ \bf{ \: f'(x) = \: x \: cosx \: + \: sinx \: \: }}}$

Additional Information :-

$\red{\boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: }}}$

$\red{\boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: }}}$

$\red{\boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1 \: }}}$

$\red{\boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga \: }}}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

f'(x) = x cosx + sinx

Step-by-step explanation:

f(x) = x sinx

so,

f'(x) = x cosx + sinx × 1

f'(x) = x cosx + sinx