Architecture

# 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.

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Notice that $lnleft(2x^{2} ight)=lnleft(2 ight)+2lnleft(x ight)$, 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 $$y”=4xe^{-y}=4x imesleft(frac{1}{2x^2} ight)=frac{2}{x}$$

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 $$equire{cancel}Big(ln(2x^2)Big)” = frac {2x}{x^2} = frac {2cancel{x}}{xcdot cancel{x}} = dfrac 2{x}$$

But avoid

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