Question

Q2. Find the derivative of y (2x-1)³/ अंडर रूट 3x+1

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt y = \dfrac{(2x-1)^3}{\sqrt{3x+1}}$

Using quotient rule to differentiate given function.

$\tt\implies y\prime = \dfrac{\sqrt{3x+1}\cdot \dfrac{d}{dx}(2x-1)^3 - (2x-1)^3 \cdot \dfrac{d}{dx}(\sqrt{3x+1})}{(\sqrt{3x+1})^2}$

Now using chain rule and power rule to differentiate.

$\tt\implies y\prime = \dfrac{\sqrt{3x+1}\cdot 3(2x-1)^2\cdot 2 - (2x-1)^3 \cdot \dfrac{1}{2\sqrt{3x+1}}\cdot 3}{3x+1}$

$\tt\implies y\prime = \dfrac{6(\sqrt{3x+1})(2x-1)^2 - (2x-1)^3 \cdot \dfrac{3}{2\sqrt{3x+1}}}{3x+1}$

$\tt\implies y\prime = \dfrac{6(3x+1)(2x-1)^2 - \dfrac{3(2x - 1) ^{3} }{2}}{ \sqrt{(3x+1)^{3} }}$

$\tt\implies y\prime = \dfrac{12(3x+1)(2x-1)^2 - 3(2x - 1) ^{3}}{ 2\sqrt{(3x+1)^{3} }}$

$\tt\implies y\prime = \dfrac{(2x - 1)^{2} (12(3x+1) - 3(2x - 1))}{ 2\sqrt{(3x+1)^{3} }}$

$\tt\implies y\prime = \dfrac{(2x - 1)^{2} (36x+12 - 6x + 3)}{ 2\sqrt{(3x+1)^{3} }}$

$\boxed{\tt\implies y\prime = \dfrac{(2x - 1)^{2} (30x + 15)}{ 2\sqrt{(3x+1)^{3} }}}$