How do you differentiate $y = {ln x}^{2}$ $y = lnx^2$ ?

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How do you differentiate $y = {ln x}^{2}$ $y = lnx^2$ ?

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Answer 1 · 2016-03-04T03:25:09

$\frac{d y}{d x} = \frac{2}{x}$ $dy/dx = 2/x$

Explanation:

Applying the chain rule, along with the derivatives $\frac{d}{d x} ln (x) = \frac{1}{x}$ $d/dx ln(x) = 1/x$ and $\frac{d}{d x} x^{2} = 2 x$ $d/dx x^2 = 2x$ , we have

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

$= \frac{1}{x^{2}} (\frac{d}{d x} x^{2})$ $=1/x^2(d/dxx^2)$

$= \frac{1}{x^{2}} (2 x)$ $=1/x^2(2x)$

$= \frac{2}{x}$ $=2/x$

Answer 2 · 2016-03-04T03:33:56

$\frac{2}{x}$ $2/x$

Explanation:

Alternatively, we can simplify $ln (x^{2}) = 2 ln (x)$ $ln(x^2)=2ln(x)$ from the outset, using the rule that $log (a^{b}) = b log (a)$ $log(a^b)=blog(a)$ .

Since $\frac{d}{d x} ln (x) = \frac{1}{x}$ $d/dxln(x)=1/x$ , we see the constant can be brought from the differentiation in $\frac{d}{d x} 2 ln (x) = 2 \frac{d}{d x} ln (x) = \frac{2}{x}$ $d/dx2ln(x)=2d/dxln(x)=2/x$ .

Answer 3 · 2016-03-04T12:19:52

$\frac{2}{x}$ $2/x$

Explanation:

Just to show the versatility of calculus, we can solve this problem through implicit differentiation.

Raise both side to the power of $e$ $e$ .

$y = ln (x^{2})$ $y=ln(x^2)$

$e^{y} = e^{ln (x^{2})}$ $e^y=e^ln(x^2)$

$e^{y} = x^{2}$ $e^y=x^2$

Differentiate both sides with respect to $x$ $x$ . The left side will require the chain rule.

$e^{y} (\frac{d y}{d x}) = 2 x$ $e^y(dy/dx)=2x$

$\frac{d y}{d x} = \frac{2 x}{e^{y}}$ $dy/dx=(2x)/e^y$

Recall that $e^{y} = x^{2}$ $e^y=x^2$ .

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

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How do you differentiate $y = {ln x}^{2}$ $y = lnx^2$ ?

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