Solving a Derivative Problem using Chain Rule and Logarithmic Differentiation

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Homework Statement



Assume the notation log(a, x) implies log base a of x, where a is a constant (since I don't know LaTeX).

PROBLEM:
If y = [log(a, x^2)]^2, determine y'.

Homework Equations



Chain Rule and Logarithmic Differentiation

The Attempt at a Solution



y' = 2(log(a, x^2)) * (1/[(x^2)lna]) * (2x) = (8log(a,x))/(xlna)

Is this the correct approach and solution?
 
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S.R said:

Homework Statement



Assume the notation log(a, x) implies log base a of x, where a is a constant (since I don't know LaTeX).

PROBLEM:
If y = [log(a, x^2)]^2, determine y'.


Homework Equations



Chain Rule and Logarithmic Differentiation


The Attempt at a Solution



y' = 2(log(a, x^2)) * (1/[(x^2)lna]) * (2x) = (8log(a,x))/(xlna)

Is this the correct approach and solution?

Your approach and answer are both correct. If I may offer an alternate approach, try using the log law ##\log_ax^2=2\log_a x## at the beginning and see how that changes the rest of the problem. It's my experience that making good use of log laws at the beginning of some calculus problems makes them a little more manageable.
 
Thanks for the response. I noticed the implementation of log laws in WolframAlpha's solution where log(a,x^2) was rewritten ln(x^2)/lna.
 
gopher_p is correct: using the log law loga x2 = 2loga x does make for easier computation.

however, if you set u = loga x2 then your equation would become y = u2

so then, y'(x) = du/dx * 2u

the only "tricky" part is finding du/dx, but as S.R mentioned, you can just use the change of base formula for the logarithm...which makes it much easier to find du/dx...

u = loga x2

= (log10 x2)/(log10 a)

= (log x2)/(log a)

...then just differentiate with respect to x to find du/dx

so yes, this is definitely the correct approach to the solution!
 
Of course, it is much simpler to first use the fact that log_a(x^2)= 2log_a(x) and then differentiate.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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