calculus high school
posted by k on .
7y^4+x^3y+x=4
using implicit differentation, how do i solve this problem. it is a textbook example but the partwhere i "use the chain rule on the first term" doesn't make sense to me because they get
d/dx 7y^4+(x^3dy/dx +y d/dx x^3)+d/dx x

I get
28y^3(dy/dx) + x^3(dy/dx) + y(3x^2) + 1 = 0
dy/dx(28y^3 + x^3) =  1  3(x^2)(y)
dy/dx = ( 1  3(x^2)(y))/(28y^3 + x^3)
I used he chain rule on the first term, the product rule on the second term, the others were routine 
thanks. you got the answer right. but i don't understand how the chain rule is used iin this problem.

in 7y^4, according to "chain rule"
exponent times front coefficient > 4(7) = 28
keep the base, reduce the exponent by 1 y^4 becomes y^3
times the derivative of the base, the base is y, so derivative of y is called dy/dx
result > 28y^3(dy/dx) 
isn't that the power rule?
anyway, ow do you derive the second term? 
never mind, the second term is derived using the product rule

Chain Rule with single variable:
d/dx [3(2x4)^2]
=6(x4)(2)=12x48
Chain Rule with multiple variables:
d/dx [3y^2]
=6y(dy/dx)
The y and the (x4) terms serve a similar function. You technically use the chain rule all the time. The derivative of a variable, like x, is simply 1. So taking the derivative of 3x^2 is really easy.
d/dx [3x^2]
=6x(1)
But (2x4) and y aren't the same as x, and neither are the derivatives. So the chain rule will produce something appreciable. The derivative of 2x4 is 2, so it affects the expression (whereas the derivative of x is simply one, and multiplying by one changes nothing). Y is also not equal to x. Because we are differentiating with respect to x, d/dx of y cant be 1. And we don't know anything else about y, so all we can do is say that dy/dx equals... well dy/dx.