Answer to Question #214141 in Calculus for sbuda

Question #214141

Differentiate the following with regard to x :

y=x^4sinx


1
Expert's answer
2021-07-06T15:39:52-0400

Given the equation


y=x4sinxy=x^{4sinx}

The derivative of the function is obtained thus:


ddx[x4sin(x)]=x4sin(x)ddx[ln(x)4sin(x)]=x4sin(x)4ddx[ln(x)sin(x)]=4x4sin(x)(ddx[ln(x)]sin(x)+ln(x)ddx[sin(x)])=4x4sin(x)(1xsin(x)+ln(x)cos(x))=4x4sin(x)(sin(x)x+cos(x)ln(x)) We can rewrite the above result as: =x4sin(x)(4sin(x)x+4cos(x)ln(x))\begin{aligned} &\begin{gathered} \quad \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{4 \sin (x)}\right] \\ =x^{4 \sin (x)} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\ln (x) \cdot 4 \sin (x)] \\ =x^{4 \sin (x)} \cdot 4 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\ln (x) \sin (x)] \\ =4 x^{4 \sin (x)}\left(\frac{\mathrm{d}}{\mathrm{d} x}[\ln (x)] \cdot \sin (x)+\ln (x) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\sin (x)]\right) \\ =4 x^{4 \sin (x)}\left(\frac{1}{x} \sin (x)+\ln (x) \cos (x)\right) \\ =4 x^{4 \sin (x)}\left(\frac{\sin (x)}{x}+\cos (x) \ln (x)\right) \end{gathered}\\ &\text { We can rewrite the above result as: }\\ &=x^{4 \sin (x)}\left(\frac{4 \sin (x)}{x}+4 \cos (x) \ln (x)\right) \end{aligned}


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