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

Answer 1

#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#
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Answer 2

#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#.
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Answer 3

#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#
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Answer 4

To differentiate ( y = \ln(x^2) ), you can use the chain rule. The chain rule states that if you have a function inside another function, you differentiate the outer function first, then multiply by the derivative of the inner function.

The derivative of ( \ln(u) ) with respect to ( u ) is ( \frac{1}{u} ), and the derivative of ( x^2 ) with respect to ( x ) is ( 2x ).

So applying the chain rule:

[ \frac{dy}{dx} = \frac{1}{x^2} \cdot 2x ]

[ \frac{dy}{dx} = \frac{2x}{x^2} ]

[ \frac{dy}{dx} = \frac{2}{x} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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