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Question
Differentiate the following w.r.t.x: `5^(sin^3x + 3)`
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Solution
Let y = `5^(sin^3x + 3)`
Differentiating w.r.t. x,we get,
`"dy"/"dx" = "d"/"dx"(5^(sin^3x + 3))`
= `5^(sin^3x + 3)·log5·"d"/"dx"(sin^3x + 3)`
= `5^(sin^3x + 3)·log5·[3sin^2x·"d"/"dx"(sin x) + 0]`
= `5^(sin^3x + 3)·log5·[3sin^2x cosx]`
= `3sin^2x cosx·5^(sin^3x + 3)·log5`.
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