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$f(x)=x^{14x}$

Find $f'(x)$

I used the chain rule and wrote it as $f(U)=U^{14x},U(x)=x $, and get an answer :$14x(x)^{14x-1}$

But it is wrong .The right answer should be make $y=x^{14x}$ then $\ln y=\ln x^{14x}$ then $\ln y=14x\ln x$ then differentiate each side with respect to $x$ . Can anyone explain why my method is wrong ?

Wolgwang
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Scarlett
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    Please use MathJax to formulate the equations. Is the function this one? $$ f(x) = x^{14x} $$ – Matti P. Nov 09 '20 at 07:51
  • yes ,sorry ,I did not know about Math Jax , first time here . – Scarlett Nov 09 '20 at 07:51
  • https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Matti P. Nov 09 '20 at 07:52
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    Derivative $U^{a}$ is $aU^{a-1}$ if $a$ is a constant. It is not true when $a$ is itself a function of $x$. – Kavi Rama Murthy Nov 09 '20 at 07:52
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    There are two rules to use. $(f^g)' = g\cdot f^{g-1} \cdot f'$ if $g$ is constant and $(f^g)' = \ln f \cdot f^g \cdot g'$ if $f$ is constant. If neither are constant, the answer is the sum of both options. – Ninad Munshi Nov 09 '20 at 07:58
  • I understand that but the chain rule can also be used on 2 functions without any constant, my outside function here is 14x and inside function here is x, so I used the chain rule. The derivative of outside function times the derivate of inside function, but it turns out to be wrong. I am confused why chain rule does not work here? – Scarlett Nov 09 '20 at 07:58
  • sorry for the format , still couldnt figure out the math jax – Scarlett Nov 09 '20 at 07:58
  • @Scarlett your mistake is in both "outside" and "inside" function. If the outside function were $f(x) = x$ and the inside $g(x) = 14x$, then $f(g(x)) = 14x$, which is not what you have. – Ninad Munshi Nov 09 '20 at 07:59
  • My hint for the problem: Perhaps you have seen a similar question about calculating the derivative of $x^x$. That's a tricky one, because it's neither an exponential function nor power function. The trick is to rewrite the function with logarithms, because $$ x^x = e^{x \ln x} $$ – Matti P. Nov 09 '20 at 08:01
  • @matti P that is helpful ,thanks. – Scarlett Nov 09 '20 at 08:07
  • @ninad munshi I was considering 14x as outside function and x as an inside function if that makes any difference. – Scarlett Nov 09 '20 at 08:08
  • @Scarlett you tell me, given the logic I have provided, does it make a difference? – Ninad Munshi Nov 09 '20 at 08:15
  • @ninad I could not follow the part f(g(x))=14x – Scarlett Nov 09 '20 at 08:17

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There are two rules to use. The first $$(f^g)' = g\cdot f^{g-1} \cdot f'$$ if $g$ is constant and $$(f^g)' = \ln f \cdot f^g \cdot g'$$ if $f$ is constant. If neither are constant, the answer is the sum of both options:

$$(f^g)' = g\cdot f^{g-1} \cdot f' + \ln f \cdot f^g \cdot g'$$

This is also chain rule, but in a different form. You might have seen this pattern in product rule:

$$(fg)' = f'g+fg'$$

where you ferret out the dependence (derivative) in one function at a time. This is the intuition you can carry forward if you are careful about it.

Ninad Munshi
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    I've never thought about the product rule or the differentiation rule for $f(x)^{g(x)}$ in this way before. Excellent instructive answer. – K.defaoite Nov 09 '20 at 11:57
  • @ninad munshi in my example , neither is constant , f is x and g is 14x , per your answer , I would use the third formula you wrote , but it would not give me the correct answer either unless I use the method log and differentiate both sides. – Scarlett Nov 09 '20 at 20:22
  • @Scarlett it would give you the correct answer. If you provide the work for your attempt, I'd be happy to point out what might have gone wrong. – Ninad Munshi Nov 09 '20 at 20:26
  • @K.defaoite yes it is really useful to realize there really is only chain rule and that it operates one at a time. All first year calculus product rule is really multivariable chain rule on the function $f(x,y) = xy$ and $x=x(t)$, $y=y(t)$ – Ninad Munshi Nov 09 '20 at 20:29
  • @ninad munshi thank you so much. I wrote it down on the paper , just did not know how to use formatting on this website. Let me try . – Scarlett Nov 09 '20 at 20:32
  • @ninad I ran into a similar but different one, I was given 5x rise to the power of 14x, find derivative . In that case, to use your third formula, I think f would be x, g would be 14x, so I pulled 5 to the front and wrote 5 (derivative of x^14x), solved it that way, the result is wrong. Do you see anything wrong in my thinking? I promise one day I will learn how to format here. – Scarlett Nov 10 '20 at 20:52
  • @Scarlett it seems like you were on the right track, except parentheses matter. $5x^{14x} \neq (5x)^{14x}$ because the one on the left is $5\cdot(x^{14x})$ but the one on the right is equivalent to $(5^{14x})\cdot (x^{14x})$ – Ninad Munshi Nov 10 '20 at 22:02
  • @ninad munshi I know what you are saying, I did use the parentheses. So I pulled the 5 to the front and put the rest in the parentheses, and solve the derivate that inside the parentheses, use that times 5 in the end. but that gave me the wrong answer. Wish I could upload my work here , I will have time to learn how to do that this weekend. Thank you. – Scarlett Nov 10 '20 at 23:57