Question

find dy/dx if y= given value​

Answer 1

Explanation:

$y = {x}^5 + {x}^{3} + 4 \sqrt{x} + 7$

Differentiate with respect to x, then

$\frac{dy}{dx} = \frac{d}{dx}( {x}^{5} + {x}^3 + 4 \sqrt{x} + 7) = > \frac{dy}{dx}$

$= 5 {x}^{4} + 3 {x}^{2} + \frac{4}{2 \sqrt{x} } + 0 = > \frac{dy}{dx} =$

$5 {x}^4 + 3 {x}^{2} + \frac{2}{ \sqrt{x} }$

Hope it helps you.

Answer 2

d(x^n) /dx = nx^n-1

and

d (constant)/dx = 0

$y = x {}^{5} + x {}^{3} + 4x {}^{ \frac{1}{2} } + 7$

then dy/dx is

$= 5 x{}^{4} + 3x {}^{2} + \frac{2}{ \sqrt{x} }$

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