Maria G.
asked • 06/03/16how would i solve dy/dx
y=x^(log3(x))
Then
dydx = ??
please express answer in terms of natural logs
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2 Answers By Expert Tutors
Mark M. answered • 06/03/16
Tutor
4.9
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Using the change of base formula, log3x = lnx/ln3
So. the given function is equivalent to y = xlnx/ln3
To find dy/dx, use logarithmic differentiation:
lny = lnxlnx/ln3
= (lnx/ln3)lnx
lny = (1/ln3)(lnx)2
y'/y = (1/ln3)[2(lnx)(1/x)]
y'/y = (2/ln3)[lnx/x]
y' = dy/dx = y(2/ln3)[lnx/x]
= xlnx/ln3(2/ln3)[lnx/x]
Christian H. answered • 06/06/16
Tutor
New to Wyzant
Mathematics Tutor
I have a different solution. My solution will not require the change of base formula. Now, if you go and check the solution that wolfram gives you might be tempted to say that my solution is wrong. It only looks that way because if you input the function that you want to differentiate Wolfram is going to decide to use the chain rule to find it's derivative which will produce an answer that's gonna look very different form the one I have. Lets get started:
1. y = xlog3(x) That's the problem you wanna differentiate right?
2. Lets take the log of base 3 of both sides:
2. Lets take the log of base 3 of both sides:
log3(y) = log3(xlog3(x) )
3. Using a property of logarithms that you learn in Precalculus:
log3(y) = log3(x) * log3(x)
4. Using another property of logarithms that you learn in Precalculus:
log3(y) = log3(x + x)
5. Simplify:
log3(y) = log3(2x)
6. Now we can differentiate:
[log3(y)]’ = [log3(2x)]'
7. [y' / yln(3) ] = [2 / (2xln(3)) ]
8. Simplify:
[y' / yln(3) ] = [1 / xln(3) ]
9. Solve for y' :
y' = y/x
10. That's it.
Mark M.
tutor
I don't follow how you got from step 3 to step 4
Report
06/06/16
Christian H.
You're totally right. sorry about that. I confused myself by thinking that multiplication of logarithms means I get to to add the arguments. I'm wrong about that and admit it. Here's what it should look like continuing from step 3:
(by the way thanks for catching that mistake, I wrote this up at 4 in the morning and I probably should not have done this so late.)
3. log3(y) = log3(x) * log3(x)
4. log3(y) = [log3(x)]2
5. Now at this point were are going to actually differentiate. We will follow the rules of implicit differentiation on the left side with y and the right side will make use of the chain rule. Ready? Here we go:
[log3(y)]' = {[log3(x)]'}2
6. Implicitly differentiating the left side will yield the same result as before:
[y'/ln(3)y]
7. Now let's differentiate the right side using the Chain Rule which is going to be a tiny (not a lot) bit of work for us:
To set up some notation that I like to use, when using the Chain Rule I like to call my inner function u, and I like to call my outer function f(u) - So here we go:
u = log3(x) f(u) = u2
u' = [1/ln(3)x] f'(u) = 2u
The chain rule says we take the derivative of the inner function, the outer function, multiply them and then we are done! So let's go ahead and multiply u' and f'(u) to get
u' * f'(u) ⇒ [1/ln(3)x] * 2u
⇒ 2u/[ln(3)x]
But What's u? u is the outer function I defined above when we started the chain rule so let's go ahead and substitute u back in.
⇒ [2log3(x)] / [ln(3)x]
and that's it for differentiating the right side! Now lets put everything together and solve for y'!
8. [y' / ln(3)y] = [2log3(x)] / [ln(3)x]
9. Solve for y':
10. ⇒ y' = [ [2log3(x)] / [ln(3)x] ] * ln(3)y
11. ⇒ cancel the ln(3) ' s and the final answer we get is:
12. ⇒ y' = 2log3(x)y
Report
06/06/16
Christian H.
To Mark M:
Does this look good to you?
Report
06/06/16
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