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प्रश्न
Differentiate w.r.t. x the function:
sin3 x + cos6 x
उत्तर
Let, y = sin3 x + cos6 x
On differentiating with respect to x,
`dy/dx = d/dx sin^3 x + d/dx cos^6 x`
`= 3 sin^2 x d/dx (sin x) + 6cos^5 x d/dx (cos x)`
`= 3 sin^2 x cos x + 6 cos^5 x (- sin x)`
`= 3 sin^2 x cos x - 6 cos^5 x sin x`
= 3 sin x cos x (sin x - 2 cos4 x)
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