Second derivatives with implicit differentiation

 
 
Second derivatives with implicit differentiation blog post.jpeg
 
 
 

Don’t forget to plug the first derivative into the second derivative

Before, when we needed to find the derivative of a function of yy in terms of xx, like y=f(x)y=f(x), we used our derivative rules to find the first derivative, and then we took the derivative of the first derivative to get the second derivative.

When we have an equation in terms of xx and yy that can’t be easily solved for yy, we can use implicit differentiation to find its first and second derivative.

Krista King Math.jpg

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

 

Remember that we’ll use implicit differentiation to take the first derivative, and then use implicit differentiation again to take the derivative of the first derivative to find the second derivative.

Once we have an equation for the second derivative, we can always make a substitution for yy, since we already found y′y' when we found the first derivative.

 
 

Using implicit differentiation to find the first and second derivatives of an implicitly-defined function

 
Krista King Math Signup.png
 
Calculus 1 course.png

Take the course

Want to learn more about Calculus 1? I have a step-by-step course for that. :)

 
 
 
 

Finding a second derivative using implicit differentiation

Example

Find the second derivative.

2y2+6x2=762y^2+6x^2=76

Because it’s a little tedious to isolate yy in this equation, we’ll use implicit differentiation to take the derivative. Remember that we have to multiple by y′y' or dy/dxdy/dx whenever we take the derivative of yy.

The first derivative is

4y⋅y′+12x=04y\cdot y'+12x=0

4yy′=−12x4yy'=-12x

y′=−12x4yy'=\frac{-12x}{4y}

y′=−3xyy'=-\frac{3x}{y}

We’re looking for the second derivative, y′′y'', of our original function, so we need to take the derivative of y′y'. Using quotient rule and implicit differentiation together, the second derivative is

y′′=−(3)(y)−(3x)(1⋅y′)(y)2y''=-\frac{(3)(y)-(3x)(1\cdot y')}{(y)^2}

y′′=−3y−3xy′y2y''=-\frac{3y-3xy'}{y^2}

Since the first derivative we found earlier gives us a value for y′y', we can plug that into the second derivative and then simplify to get our final answer.

y′′=−3y−3x(−3xy)y2y''=-\frac{3y-3x\left(-\frac{3x}{y}\right)}{y^2}

y′′=−3y+9x2yy2y''=-\frac{3y+\frac{9x^2}{y}}{y^2}

y′′=−3y2y+9x2yy2y''=-\frac{\frac{3y^2}{y}+\frac{9x^2}{y}}{y^2}

y′′=−9x2+3y2yy2y''=-\frac{\frac{9x^2+3y^2}{y}}{y^2}

y′′=−9x2+3y2y(1y2)y''=-\frac{9x^2+3y^2}{y}\left(\frac{1}{y^2}\right)

y′′=−9x2+3y2y3y''=-\frac{9x^2+3y^2}{y^3}

Second derivatives with implicit differentiation for Calculus 1.jpg

Since the first derivative we found earlier gives us a value for y', we can plug that into the second derivative and then simplify to get our final answer.

Example

Find the second derivative.

3y2x+24x3=6y43y^2x+24x^3=6y^4

We’ll use implicit differentiation, remembering to multiply by y′y' or dy/dxdy/dx whenever we take the derivative of yy. Remember to use product rule for 3y2x3y^2x.

(6y⋅dydx)(x)+(3y2)(1)+72x2=24y3⋅dydx\left(6y\cdot \frac{dy}{dx}\right)(x)+(3y^2)(1)+72x^2=24y^3\cdot \frac{dy}{dx}

6xy(dydx)+3y2+72x2=24y3(dydx)6xy\left(\frac{dy}{dx}\right)+3y^2+72x^2=24y^3\left(\frac{dy}{dx}\right)

6xy(dydx)−24y3(dydx)=−72x2−3y26xy\left(\frac{dy}{dx}\right)-24y^3\left(\frac{dy}{dx}\right)=-72x^2-3y^2

(dydx)(6xy−24y3)=−72x2−3y2\left(\frac{dy}{dx}\right)(6xy-24y^3)=-72x^2-3y^2

dydx=−72x2−3y26xy−24y3\frac{dy}{dx}=\frac{-72x^2-3y^2}{6xy-24y^3}

dydx=−24x2+y22xy−8y3\frac{dy}{dx}=-\frac{24x^2+y^2}{2xy-8y^3}

To find the second derivative, we’ll use quotient rule and implicit differentiation together.

d2ydx2=−[48x+2y(dydx)](2xy−8y3)−(24x2+y2)[((2)(y)+(2x)(1(dydx)))−24y2(dydx)](2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left[48x+2y\left(\frac{dy}{dx}\right)\right]\left(2xy-8y^3\right)-\left(24x^2+y^2\right)\left[\left((2)(y)+(2x)\left(1\left(\frac{dy}{dx}\right)\right)\right)-24y^2\left(\frac{dy}{dx}\right)\right]}{\left(2xy-8y^3\right)^2}

d2ydx2=−[48x+2y(dydx)](2xy−8y3)−(24x2+y2)[2y+2x(dydx)−24y2(dydx)](2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left[48x+2y\left(\frac{dy}{dx}\right)\right]\left(2xy-8y^3\right)-\left(24x^2+y^2\right)\left[2y+2x\left(\frac{dy}{dx}\right)-24y^2\left(\frac{dy}{dx}\right)\right]}{\left(2xy-8y^3\right)^2}

We can use our first derivative to substitute for dy/dxdy/dx.

d2ydx2=−[48x+2y(−24x2+y22xy−8y3)](2xy−8y3)−(24x2+y2)[2y+2x(−24x2+y22xy−8y3)−24y2(−24x2+y22xy−8y3)](2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left[48x+2y\left(-\frac{24x^2+y^2}{2xy-8y^3}\right)\right]\left(2xy-8y^3\right)-\left(24x^2+y^2\right)\left[2y+2x\left(-\frac{24x^2+y^2}{2xy-8y^3}\right)-24y^2\left(-\frac{24x^2+y^2}{2xy-8y^3}\right)\right]}{\left(2xy-8y^3\right)^2}

d2ydx2=−[48x−2y(24x2+y22xy−8y3)](2xy−8y3)−2(24x2+y2)[y−x(24x2+y22xy−8y3)+12y2(24x2+y22xy−8y3)](2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left[48x-2y\left(\frac{24x^2+y^2}{2xy-8y^3}\right)\right]\left(2xy-8y^3\right)-2\left(24x^2+y^2\right)\left[y-x\left(\frac{24x^2+y^2}{2xy-8y^3}\right)+12y^2\left(\frac{24x^2+y^2}{2xy-8y^3}\right)\right]}{\left(2xy-8y^3\right)^2}

This is a huge answer, but that’s not unusual when you’re dealing with complex implicit differentiation. So just try to simplify as much as you can, but don’t worry too much about getting a really clean answer.

d2ydx2=−(48x−48x2y+2y32xy−8y3)(2xy−8y3)−(48x2+2y2)(y−24x3+xy22xy−8y3+288x2y2+12y42xy−8y3)(2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left(48x-\frac{48x^2y+2y^3}{2xy-8y^3}\right)\left(2xy-8y^3\right)-\left(48x^2+2y^2\right)\left(y-\frac{24x^3+xy^2}{2xy-8y^3}+\frac{288x^2y^2+12y^4}{2xy-8y^3}\right)}{\left(2xy-8y^3\right)^2}

d2ydx2=−(48x−48x2y+2y32xy−8y3)(2xy−8y3)−(48x2+2y2)(y−24x3+xy2−288x2y2−12y42xy−8y3)(2xy−8y3)2\frac{d^2y}{dx^2}=-\frac{\left(48x-\frac{48x^2y+2y^3}{2xy-8y^3}\right)\left(2xy-8y^3\right)-\left(48x^2+2y^2\right)\left(y-\frac{24x^3+xy^2-288x^2y^2-12y^4}{2xy-8y^3}\right)}{\left(2xy-8y^3\right)^2}

d2ydx2=(48x2+2y2)(y−24x3+xy2−288x2y2−12y42xy−8y3)−(48x−48x2y+2y32xy−8y3)(2xy−8y3)(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{\left(48x^2+2y^2\right)\left(y-\frac{24x^3+xy^2-288x^2y^2-12y^4}{2xy-8y^3}\right)-\left(48x-\frac{48x^2y+2y^3}{2xy-8y^3}\right)\left(2xy-8y^3\right)}{\left(2xy-8y^3\right)^2}

d2ydx2=48x2y−48x2(24x3+xy2−288x2y2−12y4)2xy−8y3+2y3−2y2(24x3+xy2−288x2y2−12y4)2xy−8y3−[96x2y−384xy3−2xy(48x2y+2y3)2xy−8y3+8y3(48x2y+2y3)2xy−8y3](2xy−8y3)2\frac{d^2y}{dx^2}=\frac{48x^2y-\frac{48x^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}+2y^3-\frac{2y^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}-\left[96x^2y-384xy^3-\frac{2xy\left(48x^2y+2y^3\right)}{2xy-8y^3}+\frac{8y^3\left(48x^2y+2y^3\right)}{2xy-8y^3}\right]}{\left(2xy-8y^3\right)^2} 

d2ydx2=48x2y−48x2(24x3+xy2−288x2y2−12y4)2xy−8y3+2y3−2y2(24x3+xy2−288x2y2−12y4)2xy−8y3−96x2y+384xy3+2xy(48x2y+2y3)2xy−8y3−8y3(48x2y+2y3)2xy−8y3(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{48x^2y-\frac{48x^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}+2y^3-\frac{2y^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}-96x^2y+384xy^3+\frac{2xy\left(48x^2y+2y^3\right)}{2xy-8y^3}-\frac{8y^3\left(48x^2y+2y^3\right)}{2xy-8y^3}}{\left(2xy-8y^3\right)^2}

d2ydx2=2y3−48x2y+384xy3+2xy(48x2y+2y3)−8y3(48x2y+2y3)−48x2(24x3+xy2−288x2y2−12y4)−2y2(24x3+xy2−288x2y2−12y4)2xy−8y3(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{2y^3-48x^2y+384xy^3+\frac{2xy\left(48x^2y+2y^3\right)-8y^3\left(48x^2y+2y^3\right)-48x^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)-2y^2\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}}{\left(2xy-8y^3\right)^2}

d2ydx2=2y3−48x2y+384xy3+(2xy−8y3)(48x2y+2y3)−(48x2+2y2)(24x3+xy2−288x2y2−12y4)2xy−8y3(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{2y^3-48x^2y+384xy^3+\frac{\left(2xy-8y^3\right)\left(48x^2y+2y^3\right)-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}}{\left(2xy-8y^3\right)^2}

d2ydx2=(2y3−48x2y+384xy3)(2xy−8y3)2xy−8y3+(2xy−8y3)(48x2y+2y3)−(48x2+2y2)(24x3+xy2−288x2y2−12y4)2xy−8y3(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{\frac{\left(2y^3-48x^2y+384xy^3\right)\left(2xy-8y^3\right)}{2xy-8y^3}+\frac{\left(2xy-8y^3\right)\left(48x^2y+2y^3\right)-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}}{\left(2xy-8y^3\right)^2}

d2ydx2=(2y3−48x2y+384xy3)(2xy−8y3)+(2xy−8y3)(48x2y+2y3)−(48x2+2y2)(24x3+xy2−288x2y2−12y4)2xy−8y3(2xy−8y3)2\frac{d^2y}{dx^2}=\frac{\frac{\left(2y^3-48x^2y+384xy^3\right)\left(2xy-8y^3\right)+\left(2xy-8y^3\right)\left(48x^2y+2y^3\right)-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{2xy-8y^3}}{\left(2xy-8y^3\right)^2}

d2ydx2=(2y3−48x2y+384xy3)(2xy−8y3)+(2xy−8y3)(48x2y+2y3)−(48x2+2y2)(24x3+xy2−288x2y2−12y4)(2xy−8y3)3\frac{d^2y}{dx^2}=\frac{\left(2y^3-48x^2y+384xy^3\right)\left(2xy-8y^3\right)+\left(2xy-8y^3\right)\left(48x^2y+2y^3\right)-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{\left(2xy-8y^3\right)^3}

d2ydx2=(2xy−8y3)[(2y3−48x2y+384xy3)+(48x2y+2y3)]−(48x2+2y2)(24x3+xy2−288x2y2−12y4)(2xy−8y3)3\frac{d^2y}{dx^2}=\frac{\left(2xy-8y^3\right)\left[\left(2y^3-48x^2y+384xy^3\right)+\left(48x^2y+2y^3\right)\right]-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{\left(2xy-8y^3\right)^3}

d2ydx2=(2xy−8y3)(4y3+384xy3)−(48x2+2y2)(24x3+xy2−288x2y2−12y4)(2xy−8y3)3\frac{d^2y}{dx^2}=\frac{\left(2xy-8y^3\right)\left(4y^3+384xy^3\right)-\left(48x^2+2y^2\right)\left(24x^3+xy^2-288x^2y^2-12y^4\right)}{\left(2xy-8y^3\right)^3}

 
Krista King.png
 

Get access to the complete Calculus 1 course

 
Learn mathKrista Kingmath, learn online, online course, online math, equilibrium solutions, calculus 1, calculus i, calc 1, calc i, implicit differentiation, implicit derivatives, second derivatives, derivatives, differentiation