MILY

HARDER PARTS WAS 3(x^2+y^2)^2=26(y^2+y^2) Find the equation of the tangent line to the curve (a lemniscate) 3(x^2+y^2)^2=26(y^2+y^2) at the point (-4,2). The equation of this tangent line can be written in the form y=mx+b where m is:? and where b is:?

anti-derivation given f"(x)=3x+0 and f'(-1)=1 and f(-1)=5 find f'(x)=? and find f(2)

Evaluate the indefinite integral (a.)∫√(cotx)csc^2xdx (b.)∫sec^3xtanxdx

the fundamental theorem of calculus, f(x)=∫(0,x) t^3+2t^2+ 2 dt, and find f"(x). my answer was: f'(x)=x^3+x^2 f"(x)=3x^2+2x it said its wrong. i don't know how solve with interval, ∫(0,x)

ok i got result for m. but i don't know where is b.. i forget how do y=mx+b

oops my bad.. 26(x^2-y^2)

yea it is ∫ (t^3 + 2t^2 + 2) dt from 0 to x. thanks for ur help