Implicit differentiation of −2cos(x)sin(−3y)=3 ?

Question

I just...can't seem to understand how this implicit differentiation works, and I only have 8 tries to get it right.
I'd really appreciate if you could explain the steps in addition, since I'm hoping to learn the concept as much as if not more than to just get the credit for the answer.

Anonymous · Accepted Answer

This is made more complicated because you have to use the product rule.  If you use operator and parentheses you would solve this problem, but I can't overemphasize that even while I know that it will be ignored.

(-2cos(x)) * (sin(-3y)) = 3

u = -2cos(x)
du/dx = 2sin(x)

v = (sin(-3y))
dv/dx = cos(-3y) * (-3dy/dx) .........simple application of chain rule.

So (-2*cos(x) * cos(-3y) * (-3dy/dx)) + (sin(-3y) * (2*sin(x)) = 0

Subtract (sin(-3y) * (2*sin(x)) from both sides getting:

[(-2*cos(x) * cos(-3y) * (-3dy/dx))] = [-(sin(-3y) * (2*sin(x))]

Factor the dy/dx out of the left hand side of the equation:

dy/dx * [(-2*cos(x) * cos(-3y) * (-3)] = [-(sin(-3y) * (2*sin(x))]

Divide both sides by [(-2*cos(x) * cos(-3y) * (-3)]:

dy/dx = [-(sin(-3y) * (2*sin(x))] /  [(-2*cos(x) * cos(-3y) * (-3)]

multiply numerator and denominator by (-1)

dy/dx = [(sin(-3y) * (2*sin(x))] /  [(2*cos(x) * cos(-3y) * (-3)]  <----  Answer

or

. . . . . . . . .[(sin(-3y) * (2*sin(x))]
dy/dx = --------------------------------------  <--------- Answer (I prefer this form.)
. . . . . .  [(-2*cos(x) * cos(-3y) * (-3)]

.

? · Answer

only differentiate the full ingredient i think of the equation your watching is; 5x^3 + x^2*y - x * y^3 = 2 Differentiate to get 15 * x^2 * dx + 2 * x * y * dx + x^2 * dy - y^3 * dx - 3 * x * y^2 * dy = 0 divide by dx and convey at the same time like words (x^2 - 3xy^2) * (dy/dx) = (y^3 -15x^2 -2xy) dy/dx = (y^3 -15x^2 -2xy)/(x^2 - 3xy^2) and your finished

Trending News

Implicit differentiation of −2cos(x)sin(−3y)=3 ?

2 Answers