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(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.
To solve this problem, you are correct in using implicit differentiation. The chain rule allows us to differentiate a function that is composed of two or more functions, such as in this case where we have y expressed as a function of x.
Let's go step by step:
1. Start by differentiating both sides of the equation with respect to x:
d/dx(7y^4 + x^3y + x) = d/dx(4)
2. Now let's focus on the left side of the equation and break it down term by term applying the chain rule where necessary:
For the term 7y^4, we apply the chain rule because y is a function of x. The chain rule states that the derivative of the outer function multiplied by the derivative of the inner function.
So for 7y^4, we have:
d/dx(7y^4) = 28y^3 * dy/dx
3. For the term x^3y, we have two variables x and y, so we use the product rule to differentiate it. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
Using the product rule, we get:
d/dx(x^3y) = 3x^2*y + x^3 * dy/dx
4. Now let's differentiate the term x, which is straightforward:
d/dx(x) = 1
5. The right side of the equation is just a constant, so its derivative is zero:
d/dx(4) = 0
Putting it all together, we have:
28y^3 * dy/dx + 3x^2*y + x^3 * dy/dx + 1 = 0
Now, if you want to solve for dy/dx, you can rearrange the equation and isolate dy/dx terms:
28y^3 * dy/dx + x^3 * dy/dx = -3x^2*y - 1
Factor out dy/dx terms:
(28y^3 + x^3) * dy/dx = -3x^2*y - 1
Finally, solve for dy/dx:
dy/dx = (-3x^2*y - 1) / (28y^3 + x^3)
That's how you can solve the problem using implicit differentiation and the chain rule.