Question

$y = x5 + cotx$
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Answer 1

We have:

y = $x^5+\cot x$

We have to find, $\dfrac{dy}{dx} .$

Solution:

y = $x^5+\cot x$

Differentiating both sides w.r.t. x, we get

$\dfrac{dy}{dx}=\dfrac{d(x^5+\cot x)}{dx}$

⇒ $\dfrac{dy}{dx}=\dfrac{d(x^5)}{dx}+\dfrac{d(\cot x)}{dx}$

⇒ $\dfrac{dy}{dx}=5x^4-\csc^2x$

∴ $\dfrac{dy}{dx}=5x^4-\csc^2x$