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_______________
dy
Find —— for an implicit function:
dx
x²y – 3x = y³ – 3
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
Applying the product rule for the first term at the left-hand side:
![\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\ \mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%5B%5Cdfrac%7Bd%7D%7Bdx%7D%28x%5E2%29%5Ccdot%20y%2Bx%5E2%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28y%29%5Cright%5D-3%5Ccdot%201%3D3y%5E2%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D-0%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5B2x%5Ccdot%20y%2Bx%5E2%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D-3%3D3y%5E2%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%7D)
dy
Now, isolate —— in the equation above:
dx
Compute the derivative value at the point (– 1, 2):
x = – 1 and y = 2
I hope this helps. =)
Tags: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
_______________
dy
Find —— for an implicit function:
dx
x²y – 3x = y³ – 3
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
Applying the product rule for the first term at the left-hand side:
dy
Now, isolate —— in the equation above:
dx
Compute the derivative value at the point (– 1, 2):
x = – 1 and y = 2
I hope this helps. =)
Tags: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>