How do you differentiate $y = lnx^2$ ?

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Answer 1 · 2016-03-04T03:25:09

$dy/dx = 2/x$

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

$dy/dx = d/dxln(x^2)$

$=1/x^2(d/dxx^2)$

$=1/x^2(2x)$

$=2/x$

Answer 2 · 2016-03-04T03:33:56

$2/x$

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

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

Answer 3 · 2016-03-04T12:19:52

$2/x$

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

Raise both side to the power of $e$ .

$y=ln(x^2)$

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

$e^y=x^2$

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

$e^y(dy/dx)=2x$

$dy/dx=(2x)/e^y$

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

$dy/dx=(2x)/x^2=2/x$

How do you differentiate y = lnx^2y = lnx^2?