Question

sinx^m×cos^n...... differentiate
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Answer 1

Answer:

We know,

y = f(x).g(x) then,

dy/dx = f(x).dg(x)/dx + g(x).df(x)/dx

y = sin^m x.cosⁿx

differentiate with respect to x

dy/dx = sin^mx . {n.cos^(n-1)x (-sinx)} + cosⁿx { m.sin^(m -1)x (cosx) }

= - n.sin^(m+1)x .cos^(n-1)x +m.cos^(n+1)x.sin^(m-1)x

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