Question

If $y=f\left(\frac{3x+4}{5x+6}\right)$ and $f^{\prime }\left(x\right)=\tan x^2$ then $\frac{dy}{dx}$ is equal to

• A. $-2\tan {\left(\frac{3x+4}{5x+6}\right)}^2\times \frac{1}{{(5x+6)}^2}$
• B. $f\left(\frac{3\tan x^2+3}{5\tan x^2+6}\right)\tan x^2$
• C. $\tan x^2$
• D. None of these

Answer 1

Answer: (A)

$-2\tan {\left(\frac{3x+4}{5x+6}\right)}^2\times \frac{1}{{(5x+6)}^2}$

If $y=f\left(\frac{3x+4}{5x+6}\right)$ and $f^{\prime }\left(x\right)=\tan x^2$ then $\frac{dy}{dx}$is equal to

Differentiating w.r.t. x we get,

$\therefore \frac{dy}{dx}=f^{\prime }\left(\frac{3x+4}{5x+6}\right)\times \frac{3(5x+6)-5(3x+4)}{(5x+6{)}^2}$

$\therefore \frac{dy}{dx}=\tan {\left(\frac{3x+4}{5x+6}\right)}^2\times \frac{-2}{(5x+6{)}^2}$