Question

If $y=f\left(x^3\right),z=g\left(x^5\right),f^{\prime }\left(x\right)=\tan x$ and $g^{\prime }\left(x\right)=\sec x$ then the value of $\frac{dy}{dz}is-$

• A. $\frac{3}{5x^2}\frac{\tan x^3}{\sec x^5}$
• B. $\frac{5x^2}{3}\frac{\sec x^5}{\tan x^3}$
• C. $\frac{3x^2}{5}\frac{\tan x^3}{\sec x^5}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{5x^2}\frac{\tan x^3}{\sec x^5}$

$y=f\left(x^3\right),z=g\left(x^5\right)$
$\frac{dy}{dx}=f^{\prime }\left(x^3\right)\times 3x^2$ and $\frac{dz}{dx}=g^{\prime }\left(x^5\right)\times 5x^4$
$\frac{dy}{dz}=\frac{f^{\prime }\left(x^3\right)}{g^{\prime }\left(x^5\right)}\times \frac{3}{5}\frac{1}{x^2}$
$\frac{dy}{dz}=\frac{3}{5x^2}\times \frac{f^{\prime }\left(x^3\right)}{g^{\prime }\left(x^5\right)}$
$=\frac{3}{5x^2}\frac{\tan x^3}{\sec x^5}$