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differentiate \( \sin \left(3 x+y^{3}\right) \)

Ask by Bates Conner. in Nigeria
Jan 11,2025

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The derivative of \( \sin(3x + y^3) \) with respect to \( x \) is \( 3 \cos(3x + y^3) \).

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To differentiate \( \sin(3x + y^3) \) with respect to \( x \), we use the chain rule. The derivative will be: \[ \frac{d}{dx} \sin(3x + y^3) = \cos(3x + y^3) \cdot \frac{d}{dx}(3x + y^3) \] Now, we take the derivative of the inside: \[ \frac{d}{dx}(3x + y^3) = 3 + 3y^2 \frac{dy}{dx} \quad (\text{using implicit differentiation for } y) \] Putting it all together: \[ \frac{d}{dx} \sin(3x + y^3) = \cos(3x + y^3) \left(3 + 3y^2 \frac{dy}{dx}\right) \] This gives us the full differentiation of the function!

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