implicit differentiation

[cmoore]

Trying to help my daughter with her homework and its been a long time since using this stuff. Please help. THANKS!

Find y' (dy/dx)

sin x + cos y = sin x cos y


Find equation of the tangent line at given point using implicit differentiation:
x^(2/3) + y^(2/3) = 4 point-3(3^(1/2)), 1) (3^(1/2) is square root of 3)

y^2((y^2) - 4) = x^2((x^2) - 5) point0,-2)


derivatives of log funtions:
Find y'

y = (lnx)^(cos x)

y = [ln(1 + e^x)]^2

 

[Deleted member 4993]

Trying to help my daughter with her homework and its been a long time since using this stuff. Please help. THANKS!

Find y' (dy/dx)

sin x + cos y = sin x cos y


Find equation of the tangent line at given point using implicit differentiation:
x^(2/3) + y^(2/3) = 4 point-3(3^(1/2)), 1) (3^(1/2) is square root of 3)

y^2((y^2) - 4) = x^2((x^2) - 5) point0,-2)


derivatives of log funtions:
Find y'

y = (lnx)^(cos x)

y = [ln(1 + e^x)]^2

These are straight forward problems - no tricks involved. Some are tedious - nevertheless straight-forward.

Your daughter needs to show us some work, indicating exactly where she is stuck - so that we know where to begin to help her.

 

[Bahrom7893]

1)Take the derivative implicitly {remember, (d/dx)(sin(x)) = cos(x) and (d/dx)(cos(x)) = -Sin(x), also the formula for product rule ( u * v )' = ( uv' + vu' ) }, so we'll get:
Cos(x) + [ - Sin(y) * y' ] = Sin(x) * [ - Sin(y) * y' ] + Cos(y) * Cos(x)
2) Move all y' to one side, and the rest to the other:
Sin(x) * Sin(y) * y' - Sin(y) * y' = Cos(y) * Cos(x) - Cos(x)
3) Take y' out from one side and simplify the expression.

Part 2 is not that hard, just remember that (d/dx) [ln(u)] = (1/u) * (du/dx), and lnx^a = a*lnx. Note for the second problem, the ln is squared completely, so rewrite it as ln(1 + e^x) * ln(1 + e^x) and use the product rule. You could also use the chain rule: (d/dx) (x^n) = n * x^(n-1)

Hope this helps . Post a reply if you get stuck again