Question

Use logarithmic differentiation to differentiate y = x\sqrt[3]{1 + x^{2}}.

Select one:
A \frac{dy}{dx} = x \sqrt[3]{1 + x^{2}}[\frac{1}{x} + \frac{2x}{3(1 + 2x^{2})}]
B \frac{dy}{dx} = x \sqrt[3]{1 + x^{2}}[\frac{1}{x} + \frac{2x}{3(1 + x^{2})}]
C \frac{dy}{dx} = x \sqrt[3]{1 + x^{2}}[\frac{1}{3x} + \frac{2x}{3(1 + x^{2})}]
D \frac{dy}{dx} = x \sqrt[3]{1 + 2x^{2}}[\frac{1}{x} + \frac{2x}{3(1 + x^{2})}]

Answer 1

$\textbf{Given:}$

$y=x\,\sqrt[3]{1 + x^2}$

$\textbf{To find:}$

$\dfrac{dy}{dx}$

$\textbf{Solution:}$

$\text{Consider,}$

$y=x\,\sqrt[3]{1 + x^2}$

$\implies\,y=x(1 + x^2)^{\frac{1}{3}}$

$\text{Take logarithm on bothsides, we get}$

$\log\,y=\log(x(1 + x^2)^{\frac{1}{3}})$

$\log\,y=\log\,x+\log(1 + x^2)^{\frac{1}{3}}$

$\log\,y=\log\,x+\frac{1}{3}\log(1 + x^2)$

$\text{Differentiate with respect to 'x'}$

$\dfrac{1}{y}\,\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{3}(\dfrac{2x}{1+x^2})$

$\dfrac{1}{y}\,\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{2x}{3(1+x^2)}$

$\dfrac{dy}{dx}=y\,[\dfrac{1}{x}+\dfrac{2x}{3(1+x^2)}]$

$\implies\bf\dfrac{dy}{dx}=x\,\sqrt[3]{1 + x^2}\,[\dfrac{1}{x}+\dfrac{2x}{3(1+x^2)}]$

$\therefore\textbf{Option (B) is correct}$

Answer 2

Answer:

Option b

Step-by-step explanation: