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2. [tex]\int x^{2x}(1 + ln x)dx\[/tex] ... I set u = [tex]x^{2x}[/tex] but my du didn't quite work out right.

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2. [tex]\int x^{2x}(1 + ln x)dx\[/tex] ... I set u = [tex]x^{2x}[/tex] but my du didn't quite work out right.

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Dick

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Tom Mattson

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1. y = 3^[tex]^{}log_{}_{}_2^{}(t)[/tex] Find dy/dx

Are you sure? Unless t is some function of x, that derivative is identically zero.

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I meant dy/dt. :)

I will try to solve these again a little later, thanks.

I will try to solve these again a little later, thanks.

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lurflurf

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use the power rule

(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

or

[log(u^v)]'=v{[log(u)]'+log(u)[log(v)]'}

(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

or

[log(u^v)]'=v{[log(u)]'+log(u)[log(v)]'}

Last edited:

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Dick

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I wouldn't recommend you do that unless you are willing to carry those formulas around in your head for the rest of your life. You don't need them. Sorry, lurflurf. Did you really remember those, or did you derive them just now??use the power rule

(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

or

[log(u^v)]'=v{[log(u)]'+log(u)[log(v)]'}

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I understand how to get the answer now, thanks.

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lurflurf

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I really remember those.I wouldn't recommend you do that unless you are willing to carry those formulas around in your head for the rest of your life. You don't need them. Sorry, lurflurf. Did you really remember those, or did you derive them just now??

I think if you are going to remember 15-20 differentiation formulas that would be one of them.

If one recalls

(u^v)'=v*u^(v-1)*u' when v'=0 obvious since [x^a]'=a*x^(a-1)

(u^v)'=u^v*log(u)*v' when u'=0 obvious since [exp(a*x)]'=a*exp(a*x)

one knows by the chainn rule the sum generalizes to the case where neither is constant

it is actually easier to recall

(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

that two seperate equations

One must decide how many calculus formulas to carry in ones head (or written on the back of ones hand) at different points in life.

That formula is more helful than say

[log(sin(a*x))]'=a*cot(a*x)

and less useful than

c'=0

your results may very

in particular if I ever forget that formula it is easer to derive the general result when needed and apply it than to derive a special case.

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Dick

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Just simplify your expression before taking the derivative.1. y = 3^[tex]^{}log_{}_{}_2^{}(t)[/tex] Find dy/dx ... I tried using logarithmic differentiation but that didn't work.

Have you ever found the derivative of [tex]x^x[/tex] before? Knowing that could be quite helpful.2. [tex]\int x^{2x}(1 + ln x)dx\[/tex] ... I set u = [tex]x^{2x}[/tex] but my du didn't quite work out right.

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