Question

(3x-4)3/2 by applying chain rule find derivative..???​

Answer 1

The derivative of $(3x-4)^{\frac{3}{2} }$ is $\frac{9}{2}\sqrt{3x-4}$.

Given:

$(3x-4)^{\frac{3}{2} }$

To Find:

The derivative of $(3x-4)^{\frac{3}{2} }$  =?

Solution:

$(3x-4)^{\frac{3}{2} }$

Let u = 3x - 4 and

y = $(3x-4)^{\frac{3}{2} }$

So, y = $u^{\frac{3}{2} }$

$\frac{dy}{dx}= \frac{dy}{du} *\frac{du}{dx}$ ---------------(1)

$\frac{dy}{du}= \frac{du^{\frac{3}{2}} }{du}$

$\frac{dy}{du}= \frac{3}{2} u^{(\frac{3}{2}-1)}$

$\frac{dy}{du}= \frac{3}{2} u^{\frac{1}{2}}$ ---------------(2)

$\frac{du}{dx}=\frac{d(3x -4)}{dx}$

$\frac{du}{dx}=\frac{d3x}{dx} -\frac{d4}{dx}$

$\frac{du}{dx}=3$ -------------------(3)

Putting the values of equations (2) and (3) in equation (1), we get

$\frac{dy}{dx}= \frac{3}{2} u^{\frac{1}{2}}*3$

$\frac{dy}{dx}= \frac{9}{2} (3x-4)^{\frac{1}{2}}$

$\frac{dy}{dx}= \frac{9}{2} \sqrt{3x-4}$

The derivative of $(3x-4)^{\frac{3}{2} }$ is $\frac{9}{2}\sqrt{3x-4}$.

#SPJ1