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Differentiate w.r.t x:

Question:

Differentiate w.r.t $x: \cos \left(x^{3} \cdot e^{x}\right)$

 

Solution:

Let $y=\cos \left(x^{3} \cdot e^{x}\right), z=x^{3} \cdot e^{x}, m=e^{x}$ and $w=x^{3}$

Formula :

$\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}, \frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{n} \times \mathrm{x}^{\mathrm{n}-1}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$

According to the product rule of differentiation

$\mathrm{dz} / \mathrm{dx}=\mathrm{w} \times \frac{\mathrm{dm}}{\mathrm{dx}}+\mathrm{m} \times \frac{\mathrm{dw}}{\mathrm{dx}}$

$=\left[x^{3} \times\left(e^{x}\right)\right]+\left[e^{x} \times\left(3 x^{2}\right)\right]$

$=e^{x} \times\left[x^{3}+3 x^{2}\right]$

According to the chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dx}}$

$=-\sin \left(x^{3} \times e^{x}\right) \times\left\{e^{x} \times\left[x^{3}+3 x^{2}\right]\right\}$

 

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