calculus high school

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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

• calculus high school -

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

• calculus high school -

thanks. you got the answer right. but i don't understand how the chain rule is used iin this problem.

• calculus high school -

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)

• calculus high school -

isn't that the power rule?
anyway, ow do you derive the second term?

• calculus high school -

never mind, the second term is derived using the product rule

• calculus high school -

Chain Rule with single variable:
d/dx [3(2x-4)^2]
=6(x-4)(2)=12x-48

Chain Rule with multiple variables:
d/dx [3y^2]
=6y(dy/dx)

The y and the (x-4) 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 (2x-4) and y aren't the same as x, and neither are the derivatives. So the chain rule will produce something appreciable. The derivative of 2x-4 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.