Question

differentiate e^x sinx + x^n cos x​

Answer 1

Answer:

e^xsinx+x^ncosx

=e^xsinx.cosx+nx^n-1 cosx.-sinx

=e^xsinx.cosx-nx^n-1cosx.sinx

Answer 2

The differentiation is $cosx(e^x + nx^{n-1}) + sinx(e^x - x^n)$

Step-by-step explanation:

Given: The algebraic expression $e^x sinx + x^n cos x$

To Find: Differentiation of the given expression

Solution:

• Differentiation of the expression $e^x sinx + x^n cos x$

We have the expression $e^x sinx + x^n cos x$ such that its differentiation will be,

$\Rightarrow \dfrac{d}{dx} (e^x sinx + x^n cos x)$

$\Rightarrow \dfrac{d}{dx} (e^x sinx) + \dfrac{d}{dx}(x^n cos x)$

using the product rule for both terms, we have,

$\Rightarrow sinx \dfrac{d(e^x)}{dx} + e^x \dfrac{d(sinx)}{dx}+ cos x \dfrac{d(x^n)}{dx} + x^n \dfrac{d(cos x)}{dx}$

$\Rightarrow sinx(e^x) + e^x (cosx) + cosx (nx^{n-1}) + x^n (-sinx)$

Rearranging the above, you will get

$\Rightarrow cosx(e^x + nx^{n-1}) + sinx(e^x - x^n)$

Hence, the differentiation is $cosx(e^x + nx^{n-1}) + sinx(e^x - x^n)$