How do you find the derivative of $y = x {ln}^{3} x$ $y=xln^3x$ ?

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How do you find the derivative of $y = x {ln}^{3} x$ $y=xln^3x$ ?

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Answer 1 · 2017-08-19T23:15:38

$\frac{d y}{d x} = {ln}^{2} x (ln x + 3)$ $dy/dx=ln^2x(lnx+3)$

Explanation:

We need to start with the product rule. Where $u$ $u$ and $v$ $v$ are both functions, if $y = u v$ $y=uv$ , then:

$\frac{d y}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}$ $dy/dx=(du)/dxv+u(dv)/dx$

Thus, where $y = x {ln}^{3} x$ $y=xln^3x$ :

$\frac{d y}{d x} = (\frac{d}{d x} x) {ln}^{3} x + x (\frac{d}{d x} {ln}^{3} x)$ $dy/dx=(d/dxx)ln^3x+x(d/dxln^3x)$

Now we have two internal derivatives we need to figure out. The first is basic: $\frac{d}{d x} x = 1$ $d/dxx=1$ .

For the second derivative, we need the chain rule. First, note that we have a function cubed: ${ln}^{3} x = {(ln x)}^{3}$ $ln^3x=(lnx)^3$ . Through the power rule, $\frac{d}{d x} x^{3} = 3 x^{2}$ $d/dxx^3=3x^2$ , but if there were a more complex function instead of $x$ $x$ we see that $\frac{d}{d x} u^{3} = 3 u^{2} \frac{d u}{d x}$ $d/dxu^3=3u^2(du)/dx$ --that is, we still do the power rule but then multiply by the derivative of the inner function.

Thus, $\frac{d}{d x} {ln}^{3} x = 3 {ln}^{2} x (\frac{d}{d x} ln x)$ $d/dxln^3x=3ln^2x(d/dxlnx)$ .

$\frac{d y}{d x} = 1 \cdot {ln}^{3} x + x (3 {ln}^{2} x) (\frac{d}{d x} ln x)$ $dy/dx=1*ln^3x+x(3ln^2x)(d/dxlnx)$

Recall that $\frac{d}{d x} ln x = \frac{1}{x}$ $d/dxlnx=1/x$ :

$\frac{d y}{d x} = {ln}^{3} x + 3 x {ln}^{2} x (\frac{1}{x})$ $dy/dx=ln^3x+3xln^2x(1/x)$

$\frac{d y}{d x} = {ln}^{3} x + 3 {ln}^{2} x$ $dy/dx=ln^3x+3ln^2x$

If you wish:

$\frac{d y}{d x} = {ln}^{2} x (ln x + 3)$ $dy/dx=ln^2x(lnx+3)$

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How do you find the derivative of $y = x {ln}^{3} x$ $y=xln^3x$ ?

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How do you find the derivative of $y = x {ln}^{3} x$ $y=xln^3x$ ?