Derivatives

Implicit Differentiation

\displaystyle x^3+xy+y^2=1

\displaystyle \sin(xy)+\ln(x+y)=0,\quad x+y>0

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For relations like these, either variable is implicitly defined by the equation. It may be difficult to isolate a variable, and a relation need not define a single function over the whole plane. The first example can in fact be solved as a quadratic in $y$ , but implicit differentiation avoids having to do so.

Treat $y$ as a function of $x$ when differentiating. Every derivative of an expression involving $y$ must include the appropriate chain-rule factor.

\displaystyle x^3+xy+y^2=1

\displaystyle y=\frac{-x+\sqrt{x^2-4x^3+4}}2

\displaystyle y=\frac{-x-\sqrt{x^2-4x^3+4}}2

\displaystyle T(t)=0+(-3)(t-(1))

Input x1Branch

\displaystyle y=0,\quad \frac{dy}{dx}=\frac{-3x^2-y}{x+2y}=-3

QUESTION

When

\displaystyle y=y(x)

, how do you differentiate

\displaystyle y^2

with respect to

\displaystyle x

If $\displaystyle x^3 + xy + y^2 = 1$ , find $\displaystyle \dfrac{dy}{dx}$ at the point $\displaystyle (x, y) = (1, 0)$

If $\displaystyle \sin(tx^3) + \ln(t^2 + x) = 0$ , find $\displaystyle \dfrac{dx}{dt}$

Show that if $\displaystyle e^{2xy} +e^x + e^y = y$ that $\displaystyle {\dfrac{dx}{dy}} = {\dfrac{1}{{\dfrac{dy}{dx}}}}$

If $\displaystyle x^3+x=y$ , find $\displaystyle \dfrac{d^2y}{dx^2}$

The source's working instead differentiates $\displaystyle x^3+x=y^2$ Work through that second relation too, keeping the two answers separate

◀ Previous SectionChain Rule and Higher Derivatives Next Section ▶Related Rates

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Derivatives

Implicit Differentiation

\displaystyle x^3+xy+y^2=1

\displaystyle \sin(xy)+\ln(x+y)=0,\quad x+y>0

View range · larger means zoom out8

The curve continues beyond this window. Zoom out to see more branches, or zoom in to inspect their shape.

Treat $y$ as a function of $x$ when differentiating. Every derivative of an expression involving $y$ must include the appropriate chain-rule factor.

\displaystyle x^3+xy+y^2=1

\displaystyle y=\frac{-x+\sqrt{x^2-4x^3+4}}2

\displaystyle y=\frac{-x-\sqrt{x^2-4x^3+4}}2

\displaystyle T(t)=0+(-3)(t-(1))

Input x1Branch

\displaystyle y=0,\quad \frac{dy}{dx}=\frac{-3x^2-y}{x+2y}=-3

QUESTION

When

\displaystyle y=y(x)

, how do you differentiate

\displaystyle y^2

with respect to

\displaystyle x

If $\displaystyle x^3 + xy + y^2 = 1$ , find $\displaystyle \dfrac{dy}{dx}$ at the point $\displaystyle (x, y) = (1, 0)$

If $\displaystyle \sin(tx^3) + \ln(t^2 + x) = 0$ , find $\displaystyle \dfrac{dx}{dt}$

Show that if $\displaystyle e^{2xy} +e^x + e^y = y$ that $\displaystyle {\dfrac{dx}{dy}} = {\dfrac{1}{{\dfrac{dy}{dx}}}}$

If $\displaystyle x^3+x=y$ , find $\displaystyle \dfrac{d^2y}{dx^2}$

The source's working instead differentiates $\displaystyle x^3+x=y^2$ Work through that second relation too, keeping the two answers separate

◀ Previous SectionChain Rule and Higher Derivatives Next Section ▶Related Rates

Implicit Differentiation