How to Find d²y/dx² in Terms of x and y Using Implicit Differentiation
When a relationship between x and y is given implicitly—such as x² + y² = r² or sin(xy) = x + y—solving for y explicitly can be difficult or impossible. In these cases we rely on implicit differentiation to obtain the first derivative dy/dx, and then differentiate again to obtain the second derivative d²y/dx². In practice, the goal is to express d²y/dx² solely with x and y (and possibly dy/dx if it can be replaced by an expression in x and y). Below is a step‑by‑step guide, illustrated with two common examples, followed by a deeper explanation of the underlying calculus, a FAQ section, and a concise conclusion And it works..
Step‑by‑Step Procedure
-
Write the implicit equation
Start with the given relation F(x, y) = 0. -
Differentiate once (first derivative)
Apply d/dx to both sides, treating y as a function of x (i.e., use the chain rule: dy/dx appears whenever y is differentiated). Solve the resulting equation for dy/dx Practical, not theoretical.. -
Differentiate a second time
Differentiate the expression for dy/dx again with respect to x. This step often requires the product rule, quotient rule, and chain rule because dy/dx itself contains y and x Worth keeping that in mind. Nothing fancy.. -
Substitute dy/dx where possible
Replace any remaining dy/dx terms with the expression obtained in step 2. After substitution, the result should involve only x, y, and constants. -
Simplify
Combine like terms, factor if convenient, and present the final formula for d²y/dx².
Example 1: Circle x² + y² = r²
Step 1 – Implicit equation
(F(x,y)=x^{2}+y^{2}-r^{2}=0).
Step 2 – First derivative
Differentiate:
[
\frac{d}{dx}(x^{2})+\frac{d}{dx}(y^{2})=0 ;\Longrightarrow; 2x+2y\frac{dy}{dx}=0.
]
Solve for dy/dx:
[
\frac{dy}{dx}= -\frac{x}{y}.
]
Step 3 – Second derivative
Differentiate dy/dx:
[
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}!\left(-\frac{x}{y}\right).
]
Apply the quotient rule ( (u=-x,;v=y) ):
[
\frac{d}{dx}!\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}.
]
Here (u'=-1) and (v'=dy/dx). Thus
[
\frac{d^{2}y}{dx^{2}} = \frac{(-1),y - (-x),\frac{dy}{dx}}{y^{2}}
= \frac{-y + x\frac{dy}{dx}}{y^{2}}.
]
Step 4 – Substitute dy/dx
Insert (dy/dx = -x/y):
[
\frac{d^{2}y}{dx^{2}} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^{2}}
= \frac{-y - \frac{x^{2}}{y}}{y^{2}}
= \frac{-y^{2} - x^{2}}{y^{3}}.
]
Step 5 – Simplify using the original equation
Since (x^{2}+y^{2}=r^{2}), we have (-y^{2}-x^{2}=-(x^{2}+y^{2})=-r^{2}). Therefore
[
\boxed{\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{r^{2}}{y^{3}} }.
]
The second derivative is expressed purely in terms of x and y through the constant r² (which can also be written as x² + y² if desired) It's one of those things that adds up..
Example 2: Implicit Curve sin(xy) = x + y
Step 1 – Implicit equation
(F(x,y)=\sin(xy)-x-y=0) It's one of those things that adds up..
Step 2 – First derivative
Differentiate:
[
\cos(xy),\frac{d}{dx}(xy) - 1 - \frac{dy}{dx}=0.
]
Since (\frac{d}{dx}(xy)=y + x\frac{dy}{dx}), we get
[
\cos(xy)\bigl(y + x\frac{dy}{dx}\bigr) - 1 - \frac{dy}{dx}=0.
]
Collect dy/dx terms:
[
\cos(xy),y - 1 + \bigl[x\cos(xy)-1\bigr]\frac{dy}{dx}=0,
]
[
\frac{dy}{dx}= \frac{1 - y\cos(xy)}{x\cos(xy)-1}.
]
Step 3 – Second derivative
Differentiate the quotient (N/D) where (N=1 - y\cos(xy)) and (D = x\cos(xy)-1).
Using the quotient rule:
[
\frac{d^{2}y}{dx^{2}} = \frac{N'D - ND'}{D^{2}}.
]
We need (N') and (D'). Compute each carefully:
-
(N = 1 - y\cos(xy))
[ N' = -\frac{dy}{dx}\cos(xy) - y\bigl[-\sin(xy),(y + x\frac{dy}{dx})\bigr] = -\cos(xy)\frac{dy}{dx} + y\sin(xy)\bigl(y + x\frac{dy}{dx}\bigr). ] -
(D = x\cos(xy)-1)
[ D' = \cos(xy) + x\bigl[-\sin(xy),(y + x\frac{dy}{dx})\bigr] = \cos(xy) - x\sin(xy)\bigl(y + x\frac{dy}{dx}\bigr). ]
Plug (N, N', D, D') into the quotient‑rule expression, then replace every (dy/dx) with the fraction found in step 2. After algebraic simplification (which is lengthy but straightforward), the final result can be written as:
[ \boxed{\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{ \bigl[,x^{2}\cos^{2}(xy)-2x\cos(xy)+1,\bigr]\sin(xy) - \bigl[,2y\cos(xy)-x y\sin(xy),\bigr]\bigl[1-y\cos(xy)\bigr] } { \bigl[x\cos(xy)-1\bigr]^{3} }
