Differentiating compound functions

[GregA]

I have been happilly solving away a multitude of different questions until the book threw me this curve ball...10^(3x)

My first attempt was as follows: let y=u^3 and u=10^x
dy/du = 3u^2...du/d10 = x(10^(x-1))...3x(10^2x(10^(x-1)))...3x(10^(3x-1))

the answer given in the book however is (3ln10)10^3x...thing is I haven't met a question of this type (they have been of the sort..((x^3)^1/2)/ln(x-2) etc...) and so my best attempt to reach this answer so far is to say that 10^3X is equivilant to saying e^3xln10.

If y = e^u and u = 3xln10 then...
dy/du = e^u and du/dx = 3ln10 + (3x/10) giving...(3ln10+(3x/10))10^3x problem is...I have not reached the answer and I am not sure how I've gone wrong...please help!

 

[HallsofIvy]

I assume that you know that the derivative of e^x is just e^x itself. You should also have seen (although fewer people feel a need to memorize it!) that the derivative of a^x, where a is any positive number, is a^xln(a).

If you haven't learned that, then, whenever you have a variable as an exponent, try taking the logarithm of both sides:
If y= 10^3x, then ln(y)= 3x ln(10). Now differentiate both sides of that equation, remembering that the derivative of ln(x) is 1/x so the derivative of ln(y) with respect to x is
\frac{dln(y)}{dx}= \frac{1}{y}\frac{dy}{dx}
and then solve for \frac{dy}{dx}.

"If u = 3xln10 , the derivative of u is NOT 3ln10+ 3x/10! "10" is not a variable. u is just ax where a= 3ln10."

 

[GregA]

I assume that you know that the derivative of e^x is just e^x itself. You should also have seen (although fewer people feel a need to memorize it!) that the derivative of a^x, where a is any positive number, is a^xln(a).

Thanks for your reply HallsofIvy
don't remember having seen this(am away from my textbook to use the computer...If I have encountered it though I certainly haven't answered any questions that involved it.)I shall check the book later.

The bit you wrote at the bottom makes sense

if dy/ydx = 3ln10...dy/dx = (3ln(10))10^3x

(for what it's worth I have never come across this technique and so double thanks.)