I am trying to find the derivative of $f(x)$$=$$10^{x^2+2}$

Can anybody help me try to solve this through the chain rule?

I have tried using the chain rule and exponent rule I figured out that

$f(x)=10^x$ the derivative is $f'(x)=ln(10)10^x$

$g(x) = x^2+2$ the derivative is $g'(x) = 2x$

Hence my answer for this derivative of function f(x) was $f'(x) = (2x)(ln(10))(10^{x^2+2})$

however on the calculator/wolfram alpha it says that the answer is

enter image description here

I am unsure what the next steps to take to confirm which one is the right derivative.

  • $\begingroup$ $10^{x^2+2}=2^{x^2+2}5^{x^2+2}$. $\endgroup$
    – azif00
    Aug 22 '20 at 20:25
  • $\begingroup$ you mean $2^{x^2+3}$ ? $\endgroup$
    – jixubi
    Aug 22 '20 at 20:27
  • $\begingroup$ There are no more steps for you to take. These answers are the same. Note that WolframAlpha (and many scientific articles and pieces of scientific software) use $\log$ to denote $\ln$. And the $2$ in your $2x$ term got absorbed into the exponent: $$ (2x)(10^{x^2+2}) = (2x)(2^{x^2+2}\times 5^{x^2+2}) = x\left(2^{x^2+3} \times 5^{x^2+2}\right)$$ $\endgroup$
    – D Ford
    Aug 22 '20 at 20:28
  • $\begingroup$ thank you i was not sure since it was very complicated $\endgroup$
    – jixubi
    Aug 22 '20 at 20:29
  • $\begingroup$ To reach that use the $2$ which in the front of all the expression. $\endgroup$
    – azif00
    Aug 22 '20 at 20:29

You did it correctly and got the same answer as Wolfram did.

When Wolfram writes "$\log$" it means the natural log. (That's common, but confusing as hell, practice).

Wolfram wrote $2^{x^2+3}\times 5^{x^2 + 2}x \log 10$ and meant $2^{x^2+3}\times 5^{x^2 + 2}x \ln 10$ and you got $(2x)(ln(10))(10^{x^2+2})$.

what's the difference?

Actually nothing.

$2^{x^2+3}\times 5^{x^2 + 2}x \ln 10 =$

$2\cdot 2^{x^2 + 2}\times 5^{x^2 + 2} x \ln 10=$

$2(2\times 5)^{x^2 + 2} x\ln 10=$

$2\cdot 10^{x^2+2} x \ln 10$

which, order and parenthesis aside, is exactly what you got.

So you did jes' fine.

  • $\begingroup$ It's not that OP couldn't do this, he wrote this exactly in his post - he didn't understand why Wolfram's answer was different (both forms are actually equivalent, as people have pointed out in the comments) $\endgroup$ Aug 22 '20 at 22:31
  • $\begingroup$ Yeah.... I saw that later and ... answered in the comments. $\endgroup$
    – fleablood
    Aug 22 '20 at 22:46

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