# Find the derivative of this function using chain rule or product rule

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

• $10^{x^2+2}=2^{x^2+2}5^{x^2+2}$. Aug 22 '20 at 20:25
• you mean $2^{x^2+3}$ ? Aug 22 '20 at 20:27
• 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: $$(2x)(10^{x^2+2}) = (2x)(2^{x^2+2}\times 5^{x^2+2}) = x\left(2^{x^2+3} \times 5^{x^2+2}\right)$$ Aug 22 '20 at 20:28
• thank you i was not sure since it was very complicated Aug 22 '20 at 20:29
• To reach that use the $2$ which in the front of all the expression. 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.

• 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) Aug 22 '20 at 22:31
• Yeah.... I saw that later and ... answered in the comments. Aug 22 '20 at 22:46