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#