Asked by carolin
how do not understand how to find a deriviative of this function?Do i use chain rule and product rule?
r= the square root of (theta*sin*theta)
No one has answered this question yet.
r= the square root of (theta*sin*theta)
No one has answered this question yet.
Answers
Answered by
bobpursley
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.