Posted by carolin on Sunday, October 18, 2009 at 6:47pm.
ok, if f(u) exists, and u=G(x)
then
y= f(u)
dy/dx= d(f(u)/du * du/dx
but u=G(x)
so du/dx= G'
Now in practice.
r= sqrt(Theta*sinTheta)
dr/dtheta= 1/2 (1/sqrt(Theta*sinTheta))* d(theta*sinTheta)/dtheta
and d(theta*sinTheta)/dtheta= sinTheta+ theta*cosTheta.
Related Questions
calculus - how do not understand how to find a deriviative of this function?Do i...
Calculus - How do I use the chain rule to find the derivative of square root(1-x...
Math, Calculus - Find the derivative of the function. y= xcosx - sinx What's...
Math - Calculus Question. - hey can someone explain to me the relationship ...
math - how do i solve this? do i have to use both chain and product rule- how do...
Calculus - Am I right so far? G(x)=e^x sqrt(1+x^2) =e^x(1+x^2)^1/2 =e^x(1/2)(1+x...
calculus - use the definition of the deriviative to find the deriviative when: a...
Product Rule - Can someone please help me with this? f(x)=(x-2)^1/2 * (x^2-3)^3 ...
calculus - I have been asked to solve this two different ways. The first way is...
Calculus/Derivatives - In addition, can you walk me through how to get the ...
For Further Reading