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9 months, 1 week ago

# I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-10

Let $f(x)= \sin^{2}x\cdot e^{x}$  then f'(x) equals

• Option 1)

$e^{x}\sin 2x$

• Option 2)

$e^{x}(\sin 2x + \sin 2x)$

• Option 3)

$e^{x} (\cos ^{2}x)$

• Option 4)

$e^{x}$

108 Views
H Himanshu
Answered 9 months, 1 week ago

As we have learned

Product Rule for differentiation -

$\frac{d}{dx}{f(x).g(x)}=f(x).\frac{d}{dx}g(x)+g(x).\frac{d}{dx}f(x)$

$\Rightarrow \frac{d}{dx}(u.v)=u.\frac{dv}{dx}+v.\frac{du}{dx}$

- wherein

Take only one function for derivative along with other function.

$f'(x)= \frac{d}{dx}(\sin^{2} x)*e^{x}=$$\sin^{2}x* \frac{d}{dx}(e^{x}) +e^{x} \frac{d}{dx}(\sin x)^{2}$

$\Rightarrow f'(x)= \sin ^{2}x*e^{x} + e^{x} *2\sin x\cos x= e^{x}(\sin ^{2}x+\sin 2x)$

Option 1)

$e^{x}\sin 2x$

Option 2)

$e^{x}(\sin 2x + \sin 2x)$

Option 3)

$e^{x} (\cos ^{2}x)$

Option 4)

$e^{x}$

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