2) Differentiate arctan(xy)= 1+x^2y
to obtain dy/dx. Use implicit differentiation. Take the derivateve of both sides of the equation with respect to x, remembering that y is a function of x.
The derivative of arctan u is 1/(1+u^2).
1/[1 + (xy)^2]*[x*dy/dx + y]
= x^2*dy/dx + 2xy
y/[1 + (xy)^2] - 2 xy
= dy/dx*{-x/[1 + (xy)^2]+ x^2}
dy/dx = {y/[1 + (xy)^2] - 2 xy}/{-x/[1 + (xy)^2]+ x^2}
Hi, I have few problems that I really need help solving. I think I start out right but get messed up along the way and get the wrong answer...Any help is appreciated! thank you so much!
find derivative of:
1) y=arctan 2^x
2) arctan(xy)= 1+x^2y
Also...
3) find the equation of the tangent line to the curve y= 3arccos(x/2) at the point (1,ð)
2 answers
Let 2^x = u(x); du/dx = 2^x *ln2
y = arctan u(x)
dy/dx = dy/du*du/dx
= [1/(1+u^2)]*2^x*ln2
= [1/(1+2^(2x))]* 2^x * ln2
y = arctan u(x)
dy/dx = dy/du*du/dx
= [1/(1+u^2)]*2^x*ln2
= [1/(1+2^(2x))]* 2^x * ln2