Question

If y = f$\left( \frac{3x + 4}{5x + 6} \right)$ and f '(x) = tanx² then $\frac{dy}{dx}$ is equal to

• A. – 2 tan $\left( \frac{3x + 4}{5x + 6} \right)^{2}\left( \frac{1}{(5x + 6)^{2}} \right)$
• B. $f\left( \frac{3\tan x^{2} + 3}{5\tan x^{2} + 6} \right)\tan x^{2}$
• C. tanx²
• D. None

Answer 1

Answer: (A)

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

Answer 2

$\frac{dy}{dx}$= $f'\left( \frac{3x + 4}{5x + 6} \right).\left\lbrack \frac{(5x + 6)3 - (3x + 4)5}{(5x + 6)^{2}} \right\rbrack$

${\tan\left( \frac{3x + 4}{5x + 6} \right)}^{2}.\left\lbrack \frac{- 2}{(5x + 6)^{2}} \right\rbrack$