Differentiate the following function with respect to $x$ .
$(x^{2} + x + 1)^{4}$

Open in App

Solution

Let $y = (x^{2} + x + 1)^{4}$

Putting $x^{2} + x + 1 = u$ , we get,

$y = u^{4}$ and $u = x^{2} + x + 1$

$∴ \frac{d y}{d x} = 4 u^{3}$ and $\frac{d u}{d x} = 2 x + 1$

Now, $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$\frac{d y}{d x} = 4 u^{3} (2 x + 1) = 4 (x^{2} + x + 1)^{3} (2 x + 1)$

Suggest Corrections

Similar questions

Differentiate the following functions with respect to x :

$\frac{x^{2} - x + 1}{x^{2} + x + 1}$

Differentiate the following functions with respect to x :

$\frac{x^{2} + 1}{x + 1}$

^{'} x^{'}

\frac{x^{2} + 1}{x + 1}

(x^{3} + x^{2} + 1) s i n x

Differentiate the following functions with respect to x :

$(\frac{2 x - 1}{x^{2} + 1})$

Join BYJU'S Learning Program