Question

if y=√x(x^2+7) then dy/dx is

Answer 1

Answer:

{2Root[x(x/\2+7)]}/\-1 * [x/\2+7+2x]

Explanation:

d/dx (x/\n)= n*x/\n-1

(uv)'=u'v=uv'

First we have root then x(x/\2+7)

d/dx(x/\2+7)=2x

I hope that you find it useful

Answer 2

The differentiation of $y = \sqrt{x} (2x^{2} + 7)$ is $\frac{dy}{dx} = \frac{5x^{\frac{3}{2}}}{2} + \frac{7}{2\sqrt{x}}$

Step-by-step Explanation:

Given: $y = \sqrt{x} (2x^{2} + 7)$

To Find: $\frac{dy}{dx}$

Solution:

• Finding  $\frac{dy}{dx}$ for the function $y = \sqrt{x} (2x^{2} + 7)$

We have given $y = \sqrt{x} (2x^{2} + 7)$, which can be written as

$\Rightarrow y = \sqrt{x} (2x^{2}) + 7\sqrt{x}$

$\Rightarrow y = 2x^{\frac{5}{2}} + 7\sqrt{x}$

Now, differentiating the above with respect to 'x', we get;

$\Rightarrow \frac{dy}{dx} = \frac{dx^{\frac{5}{2}}}{dx} + 7 \frac{d\sqrt{x}}{dx}$

using the differentiation rule $\frac{dx^{n} }{dx} = n x^{n-1}$ and $\frac{d\sqrt{x} }{dx} = \frac{1}{2\sqrt{x} }$, we get

$\Rightarrow \frac{dy}{dx} = \frac{5x^{\frac{3}{2}}}{2} + \frac{7}{2\sqrt{x}}$

Hence, the differentiation of $y = \sqrt{x} (2x^{2} + 7)$ is $\frac{dy}{dx} = \frac{5x^{\frac{3}{2}}}{2} + \frac{7}{2\sqrt{x}}$