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Question

Question: How do you differentiate \[y = \ln {x^2}\]?...

How do you differentiate y=lnx2y = \ln {x^2}?

Explanation

Solution

We have given a function which is equal to a logarithmic of x2{x^2}. We have to find its derivative. Since the variable in the function is xx, we differentiate the function with respect to ‘xx’. Firstly, we take derivatives on both sides; the derivative of log of x\log {\text{ of }}x is 1x\dfrac{1}{x}, we use this property and differentiate. Then, we take the derivative of x2{x^2} which will be in the product, since derivative of xn{x^n} is nxn1n{x^{n - 1}}, so we use this property. Then, we simplify the result and we get the answer.

Complete step by step solution:
The given function is y=lnx2y = \ln {x^2}, we have to find its derivative.
Differentiating both sides with respect to ‘x’
ddx(y)=ddx(lnx2)\Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\ln {x^2}} \right)
Since the derivative of logx\log x is 1x\dfrac{1}{x}.
dydx=1x2ddx(x2)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}} \cdot \dfrac{d}{{dx}}\left( {{x^2}} \right)
Now derivative of xn{x^n} is nxn1dxn{x^{n - 1}}dx, so
dydx=1x22×xdxdx\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}} \cdot 2 \times x \cdot \dfrac{{dx}}{{dx}}
dydx=2xx2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{{x^2}}}
dydx=2x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{x}
So, dydx=2x\dfrac{{dy}}{{dx}} = \dfrac{2}{x}

Derivative of y=lnx2y = \ln {x^2} is 2x\dfrac{2}{x}.

Note:
In mathematics, logarithm is the inverse function to exponentiation. That means the logarithm of a given number xx is the exponent to which another fixed number, the base bb must be raised to produce that number xx. In the simplest case, the logarithm counts the number occurrences of the same factor in repeated multiplication, example: Since 1000=10×10×10=103 \Rightarrow 1000 = 10 \times 10 \times 10 = {10^3}.
The logarithm base 1010 of 10001000 is 33 or log10(1000)=3{\log _{10}}\left( {1000} \right) = 3. The logarithm of xx base bb is denoted as logb(x){\log _b}\left( x \right), or without parenthesis, logbx{\log _b}x or without explicit base logx\log x.
Generally, exponentiation allows any positive real number as base to be raised to any real power, always producing positive results. So, logb(x){\log _b}\left( x \right), for any two positive real number bb and xx, where bb is not equal to 11 is always unique real number yy.
Let us have a function y=f(x)y = f\left( x \right) of variable xx. The derivative of the function is the measure of the rate at which the value yy of the function changes with respect to the variable xx.