calculus

posted by .

Jena. Stop it, or you will be banned.


Using the chain rule, how do i solve this problem?
g(x)=(sin3x)^3
g'(x)=??


3sin(3x)^2 • 3

i think

*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question

*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question

*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question

let u=sin^2 (3x)
du= 2sin(3x)*cos(3x)*3

check that.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math - Calculus Question.

    hey can someone explain to me the relationship between the chain rule and implicit differentiation?
  2. MoRe HeLp PlEaSe!! (calc)

    Also can I get some help with this LONG, TIRESOME PROBLEM?
  3. Calculus

    How do I use the chain rule to find the derivative of square root(1-x^2) also, are there any general hints or tips for determining when the chain rule and product or quotient rule should be used?
  4. Calculus part 3 so sorry

    y = e^2xsin3x Find the derivative. So I used the product rule and I got sin3x(e^2x)+cos^3x(e^2x). But the actual answer is sin3x(2e^2x)+cos^3x(3e^2x) Where the the 2 and 3 before the e come from?
  5. Calculus

    Find the derivative of y with respect to x: y=3sin^4(2-x)^-1 y=[3sin(2-x)^-1]^4 y'=4[3sin(2-x)^-1]^3 (-3cos(2-x)^-1)(-1) -(2-x)^-2 y'=[12cos(2-x)^-1][3sin^3(2-x)^-1][2-x]^2 but the answer does not have a 3 in front of sin. What happened …
  6. calculus high school

    7y^4+x^3y+x=4 using implicit differentation, how do i solve this problem. it is a textbook example but the partwhere i "use the chain rule on the first term" doesn't make sense to me because they get d/dx 7y^4+(x^3dy/dx +y d/dx x^3)+d/dx …
  7. Calculus

    can you show me how to solve an equation using the chain rule
  8. Calculus

    Find f'(x) if f(x)=sin^3(4x) A. 4cos^3(4x) B. 3sin^2(4x)cos(4x) C. cos^3(4x) D. 12sin^2(4x)cos(4x) E. None of these I got D using the chain rule?
  9. Math (Calculus)

    Hello, Could somebody please help me with the following question?
  10. Calculus

    Solve using chain rule y=(3x^3+1)(-4x^2-3)^4 So far, I have: y'=(3x^3+1)*4(-4x^2-3)^3*(-8x)+(-4x^2-3)^4*(9x^2)

More Similar Questions