Let y=f (x^3) , z=g (x^5) , f^{'}(x)=\tan x and g^{'}(x)= \sec x. Then \frac{d y}{d z} is equal to

Correct Answer is : 35x2⋅tanx3 secx5

\frac{3}{5 x^2} \cdot \frac{\tan x^3}{ \sec x^5}

\frac{5 x^2}{3} \cdot \frac{ \sec x^5}{\tan x^3}

\frac{3 x^2}{5} \cdot \frac{\tan x^3}{ \sec x^5}

\frac{5}{3 x^2} \cdot \frac{ \sec x^5}{\tan x^3}

CUSAT 2022 BTech Solved Paper Shift 3

Question : 129 of 200

Marks: +1, -0

Let y=f(x3),z=g(x5),f′(x)=tan‌x and g′(x)= secx. Then ‌

is equal to

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