Question

$\huge\red{\boxed{\sf QuEsTioN}}$

Evaluate the derivative of f(x) = sin²x using Leibnitz product rule.​

Answer 1

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

The given function is

$\red{\rm :\longmapsto\:f(x) = {sin}^{2}x}$

We have to find f'(x) using Leibnitz Rule.

We know,

Leibntiz Rule for differentiation

Let us consider two function 'u' & 'v' then nth derivative of u. v is given as

$\boxed{\begin{gathered} \sf \dfrac{d^{n}}{dx}u.v = \: ^{n}C_{0} \dfrac{d^{n}u}{dx} (v) + \: ^{n}C_{1} \dfrac{d^{n - 1}u}{dx} \left(\dfrac{dv}{dx} \right) + - - + \: ^{n}C_{n}(u) \dfrac{d^{n}v}{dx} \end{gathered}}$

According to statement,

We have to find f'(x) for f(x) = sin²x

So,

$\red{\rm :\longmapsto\:f(x) = {sin}^{2}x}$

can be rewritten as

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

So, On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx} sinx. \: sinx}$

$\rm :\longmapsto\:f'(x) = \: ^{1}C_{0} \: (sinx)\dfrac{d}{dx}sinx \: + \: ^{1}C_{1} \: (sinx) \: \dfrac{d}{dx}sinx$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}sinx = cosx}$

$\boxed{ \rm \:^{n}C_{0} = 1}$

and

$\boxed{ \rm \:^{n}C_{1} = n}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = 1(sinx)(cosx) + 1(sinx)(cosx)$

$\rm :\longmapsto\:f'(x) = sinx \: cosx + sinx \: cosx$

$\bf\implies \:\boxed{ \bf \:f'(x) \: = \: 2 \: sinx \: cosx \: }$

Additional Information :-

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

Answer 2

$\huge\red{\boxed{\sf AnSwEr}}$

✰ Ǫᴜᴇsᴛɪᴏɴ

• Evaluate the derivative of f(x) = sin²x using Leibnitz product rule.

✰ ᴀɴsᴡᴇʀ

❅ Given info ≈ f(x) = sin2x

♛ Let ➜ y = sin2x

Then according to the question ,

❍ We have to use Leibnitz product rule ,

Then,

➽$\frac{dy}{dx} = ( \frac{d}{dx} )sin2x$

➜$\frac{dy}{dx} = ( \frac{d}{dx} )(sin \: x)(sin \: x)$

➜$\frac{dy}{dx} = (sin \: x)(sin \: x) + (sin \: x)(sin \: x)$

➜$\frac{dy}{dx} \: = cos \: x \: sin \: x + sin \: x \: cos \: x$

➜ $\frac{dy}{dx} = 2sin \: x \: cos \: x$

➜$\frac{dy}{dx} = \: sin2x$

✧ Therefore the answer is sin2x .