Y(x)= x^2(y) - 2cos(pix) = ln(x) - y^3
a) find the value of the derivative of y with respect to x at the point (1, -1)

Question

Y(x)= x^2(y) - 2cos(pix) = ln(x) - y^3
a) find the value of the derivative of y with respect to x at the point (1, -1)

anonymous · Answer

i know you use implicit differentiation to solve the derivative, but Idk what to do with the coordinates.

anonymous · Answer

@FibonacciChick666 @DanJS

anonymous · Answer

@solomonzelman

solomonzelman · Answer

you want the slope of the tangent line at (1,-1) and just left some text out?

solomonzelman · Answer

I have a trouble reading your Y(x), and what does that mean, y times x, or Y the function of x, and why is there 2 equal signs.

anonymous · Answer

thats how they presented the problem :/ and its y as the function of x

anonymous · Answer

for there derivative i got --> x^2 (dy/dx) + 2xy+2pisin(pix) = (1/x) -3y^2 (dy/dx)

anonymous · Answer

the* … i just need help with applying the coordinates

freckles · Answer

so based on your derivative 
and the fact you don't have three signs means you didn't mean this
"Y(x)= x^2(y) - 2cos(pix) = ln(x) - y^3"
but you meant 
"x^2(y) - 2cos(pix) = ln(x) - y^3"

freckles · Answer

so you could solve for dy/dx
and plug in (x,y)=(1,-1)