Question

y=(x^2 −1) ^4 ×log(sinx)​

Answer 1

Answer:

dxdy​=[log(sinx)]2log(sinx).tanxsec2x​−logtanx.sinxcosx​​. Now (dxdy​)π/4​=[log(sinπ/4)]2log(sinπ/4)tanπ/4sec2π/4​−logtanπ/4.sinπ/4cosπ/4​​.

Step-by-step explanation:

Answer 2

$y = ( { {x }^{2} - 1 })^{4} log(sinx) \\ \\ \frac{dy}{dx} = ( { {x}^{2} - 1 })^{4} \frac{cosx}{sinx} + log(sinx)4( { {x}^{2} - 1 })^{3} 2x \\ \\ \frac{dy}{dx} = ( { {x}^{2} - 1})^{4} cotx + 8x( { {x}^{2} - 1 })^{3} log(sinx)$