PLEASE HELP , I DON'T UNDERSTAND
find dy/dx by implicit differentiation
a) xy + x^5y^2 + 3x^3 - 4 = 1
b) lny = cos x
thank you
normally you just have an equation like
y = 3x^2
and you take the derivative of both sides, to get
y' = 6x
Then you advance to using the chain rule and product rule, and you have something like
y = e^x*sin^2(6x)
and you take the derivative of both sides to get
y' = e^x(sin^2(6x)+12sin(6x)cos(6x))
implicit differentiation just uses these techniques. Take the derivative of both sides of the equation, using the chain rule, and remembering that dx/dx = x' = 1 and dy/dx = y'
So, for your problems,
a) the derivative of xy is
d/dx(xy) = x'y + xy' = y + xy'
d/dx(x^5y^2) = (5x^4)(y^2) + (x^5)(2yy')
d/dx(3x^3) = 9x^2
d/dx(-4) = 0
d/dx(1) = 0
and put it all together to get
y + xy' + 5x^4y^2 + 2x^5yy' + 9x^2 = 0
Now just solve for y' by collecting terms
y'(x+2x^5y) + (y + 5x^4y^2 + 9x^2) = 0
y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)
b) again, use the chain rule:
d/dx(ln y) = d/dy(ln y)*dy/dx = 1/y y'
1/y y' = -sin x
y' = -y sin x
thank for you so much steve!!!
i just wanted to double check.. for (a) it is y' = - (y+5x^4y^2+9x^2)/(x+2x^5y)
i was wondering if the last part of it (x+2x^5y) is it 2x^(5y) or 2x^5(y). im not sure if the "y" is in the exponent or not.
No, since we took the derivative of x^5*y^2 we get (5x^4)(y^2) + (x^5)(2yy')
That is, 2(x^5)(y)
ok so the final answer would b
a) y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)
and
b) y' = -y sin x
thank you SO much once again steve, my exam is in a couple of hours and this solution really helped me, im very thankful. thanks so much!!
Sure! I'd be happy to help you with implicit differentiation.
Implicit differentiation is used to find the derivative of a relationship where the dependent variable is not explicitly expressed in terms of the independent variable. In other words, the equation relates x and y, but it is difficult or impossible to solve explicitly for y in terms of x.
a) To find dy/dx for the equation xy + x^5y^2 + 3x^3 - 4 = 1, you can follow these steps:
1. Start by taking the derivative of both sides with respect to x.
2. Apply the product rule to the first term, xy, to get y + x(dy/dx).
3. Apply the product rule again to the second term, x^5y^2, to get 2x^5yy' + x^5y^2(dy/dx).
4. The derivative of 3x^3 with respect to x is simply 3(3x^2) = 9x^2.
5. The derivative of 1 with respect to x is 0 (since it's a constant).
6. Simplify the equation by combining like terms.
- The derivative of x(dy/dx) + 2x^5yy' is x(dy/dx) + 2x^5yy'.
- Combine this with the remaining derivative terms: y + x(dy/dx) + x^5y^2(dy/dx) + 9x^2.
- Set this expression equal to 0 (since the derivative of a constant is 0).
7. Now, solve for dy/dx by factoring out the dy/dx term and then dividing both sides by the remaining x term:
- x(dy/dx) + 2x^5yy' + x^5y^2(dy/dx) + y + 9x^2 = 0
- Combine the dy/dx terms: x(dy/dx) + 2x^5yy' + x^5y^2(dy/dx) = - y - 9x^2
- Factor out dy/dx: [x + x^5y^2](dy/dx) + 2x^5yy' = - y - 9x^2
- Divide both sides by [x + x^5y^2]: dy/dx = (- y - 9x^2) / (x + x^5y^2)
And there you have it! The derivative dy/dx for the equation xy + x^5y^2 + 3x^3 - 4 = 1 is (- y - 9x^2) / (x + x^5y^2).
b) To find dy/dx for the equation lny = cos(x), you can follow these steps:
1. Start by taking the derivative of both sides with respect to x.
2. The derivative of lny with respect to x requires using the chain rule. The derivative of lny is (1/y) * y', where y' represents the derivative of y with respect to x.
3. The derivative of cos(x) with respect to x is -sin(x).
4. Set these derivatives equal to each other: (1/y) * y' = -sin(x).
5. Multiply both sides by y to eliminate the fraction: y' = -y * sin(x).
And there you have it! The derivative dy/dx for the equation lny = cos(x) is y' = -y * sin(x).
I hope this explanation helps! If you have any further questions, please let me know.