Question

dy/dx =x(2cos x+ xsin x )/cos ^2 x​

Answer 1

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 u got it.. nd don't forget to upvote

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Answer 2

Answer:

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 u got it.. nd don't forget to upvote