Question

value of d/dx{(sinx)^cosx} ?​

Answer 1

Answer:

Step-by-step explanation:

Y = (sin x)^cos x ,

Taking log both sides ,

LnY = cos x Ln(sin x) ,

Differentiating both sides we get ,

1/Y (dY/dx) = cos x (cos x/sin x) + (-sin x) Ln(sinx) ,

dY/dx = Y {cos x (cos x/sin x) + (-sin x) Ln(sinx)} ,

Now put Y = (sin x)^cos x ,

d/dx [(sin x)^cos x] = {(sin x)^cos x} ,

{cos x (cos x/sin x) + (-sin x) Ln(sinx)}.

Hope it helps.

Answer 2

hope this helps u. .......

step-by-step explanation:

Y = (sin x)^cos x ,

Taking log both sides ,

LnY = cos x Ln(sin x) ,

Differentiating both sides we get ,

1/Y (dY/dx) = cos x (cos x/sin x) + (-sin

x) Ln(sinx) ,

dY/dx = Y {cos x (cos x/sin x) + (-sin x)

Ln(sinx)} ,

Now put Y = (sin x)^cos x ,

d/dx [(sin x)^cos x] = {(sin x)^cos x} ,

{cos x (cos x/sin x) + (-sin x) Ln(sinx)}.

Hope it helps.