Question

Diffentiate (sinx)^cosx+(cosx)^tanx with respect to x

Answer 1

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)