Question

differentiate the following functions w.r.t x . (i) (2x^3/2-3.x^4/3-5)^5/2​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow y'=\frac{\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(15.\sqrt{x}-20.\sqrt[3]{x}\right)}{2}$

Step-by-step explanation:

Given :

$\displaystyle \sf y= \left( 2.x^{3/2}-3.x^{4/3}-5\right)^{5/2}$

We are asked to find y' :

Diff. w.r.t x :

$\displaystyle \sf \longrightarrow y'=\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{5/2-1}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)'$

$\displaystyle \sf \longrightarrow y'=\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(\left( 2.x^{3/2}\right)'-\left(3.x^{4/3}\right)'-\left(5\right)'\right)$

$\displaystyle \sf \longrightarrow y'=\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(\left( 2.x^{3/2}\right)'-\left(3.x^{4/3}\right)'-0\right)$

$\displaystyle \sf \longrightarrow y'=\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(\left( 2.x^{3/2}\right)'-\left(3.x^{4/3}\right)'\right)$

$\displaystyle \sf \longrightarrow y'=\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(\left( 2.x^{3/2}\right)'-\left(4.x^{4/3-1}\right)\right)$

$\displaystyle \sf \longrightarrow y'=\left[\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left\{\left( 3.x^{3/2-1}\right)-\left(4.x^{4/3-1}\right)\right\} \right]$

$\displaystyle \sf \longrightarrow y'=\left[\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left\{\left( 3.x^{1/2}\right)-\left(4.x^{1/3}\right)\right\} \right]$

$\displaystyle \sf \longrightarrow y'=\left[\frac{5}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left\{\left( 3.\sqrt{x}\right)-\left(4.\sqrt[3]{x}\right)\right\} \right]$

$\displaystyle \sf \longrightarrow y'=\left[\frac{1}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.5\left(3.\sqrt{x}-4.\sqrt[3]{x}\right) \right]$

$\displaystyle \sf \longrightarrow y'=\left[\frac{1}{2}.\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(15.\sqrt{x}-20.\sqrt[3]{x}\right) \right]$

$\displaystyle \sf \longrightarrow y'=\frac{\left( 2.x^{3/2}-3.x^{4/3}-5\right)^{3/2}.\left(15.\sqrt{x}-20.\sqrt[3]{x}\right)}{2}$

Hence we get required answer!