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Derivative of cubed root of x with limits  TOPIC_SOLVED

I am having trouble finding the derivative of cubed root of x using only limits. I know it is easy to use chain rule but I cannot use that yet.
by Andromeda
on Mon Oct 25, 2010 4:59 pm
 
Forum: Calculus
Topic: Derivative of cubed root of x with limits
Replies: 5
Views: 6348

Re: Derivative of cubed root of x with limits  TOPIC_SOLVED

I'm sorry, I meant the cube root of x. as in cuberoot(x).

So far a tried finding the derivative with a limit as h approaches 0
lim h-->0[(cuberoot(a+h) - cuberoot(a)) / ((a+h) - a)]

With this, I always end up with an indeterminate form.
by Andromeda
on Tue Oct 26, 2010 12:02 am
 
Forum: Calculus
Topic: Derivative of cubed root of x with limits
Replies: 5
Views: 6348

Re: Derivative of cubed root of x with limits  TOPIC_SOLVED

I'm sorry, I meant the cube root of x. as in cuberoot(x). So far a tried finding the derivative with a limit as h approaches 0 lim h-->0[(cuberoot(a+h) - cuberoot(a)) / ((a+h) - a)] With this, I always end up with an indeterminate form. \lim_{h\to0}\frac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h} use the fact t...
by Andromeda
on Tue Oct 26, 2010 4:37 pm
 
Forum: Calculus
Topic: Derivative of cubed root of x with limits
Replies: 5
Views: 6348

Trig derivatives problem  TOPIC_SOLVED

Find constants A and B such that the function y = Asin(x) + Bcos(x) satisfies the differential equation y" + y' - 2y = sin(x)

I can find the derivatives just fine, but I'm not sure where to go in order to solve for A or B.
by Andromeda
on Tue Nov 16, 2010 1:38 am
 
Forum: Calculus
Topic: Trig derivatives problem
Replies: 3
Views: 1870

Re: Trig derivatives problem  TOPIC_SOLVED

y'(x) = Acosx - Bsinx
y"(x) = -Asinx - Bcosx

(-Asinx - Bcosx) + (Acosx - Bsinx) - 2(Asinx + Bcosx) = sinx

From here I'm not sure how to solve for A or B since there are 2 unknowns and only 1 equation :confused:
by Andromeda
on Tue Nov 16, 2010 6:36 pm
 
Forum: Calculus
Topic: Trig derivatives problem
Replies: 3
Views: 1870

Implicit differentiation problem  TOPIC_SOLVED

Find y" by implicit differentiation x^3 + y^3 = 1 Here's my work: 3x^2 + 3y^2y' = 0 y' = (-3x^2)/(3y^2) = (-x^2)/(y^2) y" = (-2xy^2 + 2x^2y(dy/dx)) / y^4 = 2(-xy - x^4y^-2) / y^3 The back of the book says that the answer is -2x / y^5. I don't understand how they're simplifying it to that. ...
by Andromeda
on Tue Nov 23, 2010 6:37 pm
 
Forum: Calculus
Topic: Implicit differentiation problem
Replies: 2
Views: 2867

Re: Implicit differentiation problem  TOPIC_SOLVED

Ah I see. Thank you.
by Andromeda
on Wed Nov 24, 2010 1:22 am
 
Forum: Calculus
Topic: Implicit differentiation problem
Replies: 2
Views: 2867

Trig integral help  TOPIC_SOLVED

I am having difficulty finding integral((tanx)^2 / (secx)^2)dx.

I tried converting everything in terms of sinx and cosx but that doesn't seem to help.
by Andromeda
on Sat Jan 15, 2011 6:13 pm
 
Forum: Calculus
Topic: Trig integral help
Replies: 3
Views: 1540

Re:  TOPIC_SOLVED

I am having difficulty finding integral((tanx)^2 / (secx)^2)dx. I tried converting everything in terms of sinx and cosx but that doesn't seem to help. Once you've simplified and gotten an expression in just sine, you can use the double-angle formula (in reverse) for the cosine: . . . . . \sin^2(...
by Andromeda
on Sun Jan 16, 2011 8:50 pm
 
Forum: Calculus
Topic: Trig integral help
Replies: 3
Views: 1540

Complete the square?  TOPIC_SOLVED

I know how to complete the square to solve for roots of a non-factorable quadratic, but I don't understand the way my math book uses it to simplifiy non-factorable quadratics.. for example, the book states that "To integrate the function we complete the square in the denominator 4x^2 - 4x + 3 =...
by Andromeda
on Sat Jan 22, 2011 6:45 pm
 
Forum: Intermediate Algebra
Topic: Complete the square?
Replies: 1
Views: 1477
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