Find dy/dx
6. y^3=x^2+1
Im having trouble understanding dy/dx problems.
Us ethe chain rule.
3y^2 y' = 2x
y' = 2x/3y^2
Or, if you want it explicitly,
y = (x^2+1)^(1/3)
y' = 1/3 (x^2+1)^(-2/3) * 2x
you can see that the two are the same.
I feel your pain, buddy, but I don't know calculus
take d/dx of both sides
3 y^2 dy/dx = 2 x + 0 ( because dx/dx is 1)
dy/dx = (2/3) x/y^2
or
dy/dx = (2/3) x/ (x^2+1)^(2/3)
To find dy/dx, you need to differentiate both sides of the given equation with respect to x.
Let's start by differentiating the equation y^3 = x^2 + 1 with respect to x.
To do that, we'll use the chain rule, which states that if y = f(u), where u = g(x), then dy/dx = f'(u) * g'(x).
In this case, we can rewrite the equation as y = (x^2 + 1)^(1/3).
Now, let's find dy/dx by applying the chain rule.
Step 1: Find the derivative of the outer function f(u), which is (u)^(1/3).
- The derivative of (u)^(1/3) is (1/3) * (u)^(-2/3).
Step 2: Find the derivative of the inner function u = x^2 + 1.
- The derivative of x^2 + 1 with respect to x is 2x.
Step 3: Multiply the derivatives obtained in steps 1 and 2:
- (1/3) * (u)^(-2/3) * 2x = (2x) / (3(u)^(2/3)).
Step 4: Substitute the value of u back into the equation:
- (2x) / (3((x^2 + 1)^(2/3))).
Therefore, dy/dx = (2x) / (3((x^2 + 1)^(2/3))).