Question

find the derivative of the function:
y=
$\sqrt[3]{4x {}^{2} + 2x - 7}$


​

Answer 1

Answer:

$\frac{8x+2}{3(4x^2+2x-7)^{2/3} }$

Step-by-step explanation:

(4x^2 + 2x-7) ^ (1/3)

1/3(4x^2 + 2x-7) ^ (-2/3) * (8x + 2)

$\frac{8x+2}{3(4x^2+2x-7)^{2/3} }$

Answer 2

We have ti find derivative of the function :-

$\rm \implies \: y = \sqrt[3]{4{x }^{2} + 2x - 7}$

Differentiating both sides :-

$\rm \implies \: \dfrac{dy}{dx} =\dfrac{d(\sqrt[3]{4{x }^{2} + 2x - 7})}{dx}$

$\sf we \: can \: write \: \sqrt[3]{4{x }^{2} + 2x - 7 } \: as \: {(4{x }^{2} + 2x - 7)}^{ \frac{ 1}{3} }$

$\rm \implies \: \dfrac{dy}{dx} =\dfrac{d{({(4{x }^{2} + 2x - 7)}^{ \frac{ 1}{3} } })}{dx}$

$\rm \implies \: \dfrac{dy}{dx} = \dfrac{1}{3} \times{(4{x }^{2} + 2x - 7)}^{ \frac{ - 2}{3} } \times \dfrac{d{({4{x }^{2} + 2x - 7)} }}{dx}$

$\rm \implies \: \dfrac{dy}{dx} = \dfrac{1}{3} \times{(4{x }^{2} + 2x - 7)}^{ \frac{ - 2}{3} } \times (8x + 2)$

$\rm \implies \: \dfrac{dy}{dx} = \dfrac{ 8x + 2}{3{(4{x }^{2} + 2x - 7)}^{ \frac{ 2}{3} }}$

Answer :

$\bf \: \dfrac{d(\sqrt[3]{4{x }^{2} + 2x - 7})}{dx} = \dfrac{ 8x + 2}{3{(4{x }^{2} + 2x - 7)}^{ \frac{ 2}{3} }}$

__________________

Learn More :-

d(e^x)/dx = e^x

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

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x