Diffentiate (sinx)^cosx+(cosx)^tanx with respect to x
Answers
Answered by
0
Answer:
Given y=〖sinx〗^tanx
∴ logy=tanx logsinx .
Differentiating w.r.t. ‘x’
1/y dy/dx=d/dx (tanx logsinx)
=tanx.cosx/sinx+logsinx. sec^2x
=1+sec^2x.logsinx
dy/dx=y(1+sec^2x.logsinx).
Or, dy/dx=〖sinx 〗^tanx (1+sec^2x.logsinx).
360 views
Related Questions (More Answers Below)
Similar questions