What Is The Derivative Of Ln2X, Derivative Of Ln(2X)

In finding the derivative of $ln(2x^2),;$ I have applied the chain rule and obtained $2x / x^2.;$ Is this correct?

If not could some please explain how to do it?



If you use rules of logarithms, you don”t even have to appeal to the chain rule.

You are watching: What is the derivative of ln2x

Notice that $lnleft(2x^{2}
ight)$, so the derivative is $frac{2}{x}$.


Another possible way is: if $x
eq 0$, then by setting $y=ln(2x^2)$ we have $exp(y)=2x^2$ so $y”e^y= (e^y)”=4x$ so $$y”=4xe^{-y}=4x imesleft(frac{1}{2x^2}


Using the chain rule, if you have a function written as $h(x) = f(g(x))$, its derivative is $f”(g(x))g”(x)$. For $h(x)=ln(2x^2)$, you can take $f$ and $g$ to be $$f(x) = ln(x)\ g(x) = 2x^2$$

Deriving them gives $f”(x) = 1/x$ and $g”(x) = 4x$. Insert this into the chain rule and you get$$h”(x) = frac{1}{2x^2}cdot 4x = frac{2}{x}$$


You did just fine, and you correctly applied the chain rule. You correctly found the derivative. All you may want to do now is to simplify $$
equire{cancel}Big(ln(2x^2)Big)” = frac {2x}{x^2} = frac {2cancel{x}}{xcdot cancel{x}} = dfrac 2{x}$$

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