Question:
Differentiate w.r.t. \(x\) the function: \(sin^3x+cos^6x\)
Differentiate w.r.t. \(x\) the function: \(sin^3x+cos^6x\)
Updated On: Oct 19, 2023
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Solution and Explanation
The correct answer is \(=3sinxcosx(sinx-2cos^4x)\)
Let \(y=sin^3x+cos^6x\)
\(∴\frac{dy}{dx}=\frac{d}{dx}(sin^3x)+\frac{d}{dx}(cos^6x)\)
\(=3sin^2x.\frac{d}{dx}(sinx)+6cos^5x.\frac{d}{dx}(cosx)\)
\(=3sin^2x.cosx+6cos^5x.(-sinx)\)
\(=3sinxcosx(sinx-2cos^4x)\)
Let \(y=sin^3x+cos^6x\)
\(∴\frac{dy}{dx}=\frac{d}{dx}(sin^3x)+\frac{d}{dx}(cos^6x)\)
\(=3sin^2x.\frac{d}{dx}(sinx)+6cos^5x.\frac{d}{dx}(cosx)\)
\(=3sin^2x.cosx+6cos^5x.(-sinx)\)
\(=3sinxcosx(sinx-2cos^4x)\)
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