derivatives and differnetiation

[whiteti]

Find every point (x,y) at which the tangent line to the graph f(x)+x⁴(lnx)² is horizontal.

I do not even know where to begin, please help.

 

[serds]

Is that f(x)=x⁴(lnx)² ?

 

[whiteti]

whoops

yes i mean = not +

 

[serds]

find the point where the differentiation is 0

 

[whiteti]

?

woah, where did the u and v's come from and what do they mean?

 

[DrPhil]

Find every point (x,y) at which the tangent line to the graph f(x) = x⁴(lnx)² is horizontal.

I do not even know where to begin, please help.

Begin by differentiating, using product, power, and logarithm rules. Since the derivative is equal to the slope of the tangent, setting the derivative to zero will give you all values of x at which the tangent is horizontal.

Please show us your work so we can see where you are getting stuck.

 

[serds]

product rule?

 

[whiteti]

Begin by differentiating

I do not understand how to differentiate. I do not understand how to begin, this is what I wish to learn.

 

[JeffM]

I do not understand how to differentiate. I do not understand how to begin, this is what I wish to learn.

We are not equipped to teach an entire course in calculus. In your other thread, you asked a question about Newton's difference quotient. The derivative is the limit of such a quotient. Differential calculus teaches (among other things) rules (like the product rule) for finding the derivative of a function without going through finding the limit of the difference quotient. The derivative also equals the slope of the tangent to the curve representing a function. I do not understand how this question can be answered in a meaningful way if you do not know these things. It looks like a question from chapter two or three of a calculus text, and you seem not to have read chapter one.

 

[whiteti]

ok

so this is where im at
f'(x)=(x^4(lnx)^2)'
=(x^4)'(lnx)^2 + (x^4)(lnx)^2 '
=(4x^3)(lnx)^2 + (x^4)((lnx)^2)'

now I'm stuck...

 

[DrPhil]

Find every point (x,y) at which the tangent line to the graph f(x)=x⁴(lnx)² is horizontal.

so this is where im at
f'(x) = [x^4 (lnx)^2]'
.......= (x^4)' (lnx)^2 + (x^4) [(lnx)^2]'
.......= (4x^3) (lnx)^2 + (x^4) (???)

now I'm stuck...

ok - where you are stuck is finding the derivative of [(lnx)^2]
apply power law, then chain rule
[(lnx)^2]' = 2 (lnx) * (lnx)' = 2 (lnx) / x
because the derivative of lnx is 1/x. Then

f'(x) = (4x^3) (lnx)^2 + (x^4) (2 lnx / x))
.......= (4x^3) (lnx)^2 + (x^3)(2 lnx)
.......= 2 x^3 (lnx) (2 lnx + 1)

If any factor of f'(x) = 0, then the tangent is horizontal. I see three values to consider - but one of them may not be in the domain of f(x)
because ln(x<0) does not exist. BUT on the other hand, (x^2 lnx) may be ok at x=0 because the limit as x->0+ of (x^2 lnx) is finite??