3 Y 2 7xy 4

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Implicit Differentiation

When 1 writes a formula for a function, say $f(x) = 3x^2 + 5x + 1$, one is said to accept defined the function explicitly. Similarly, when one writes $y = 3x^2 + 5x + 1$, we have explicitly defined $y$ in terms of $x$.

All the same, for some equations relating $10$ and $y$, like $x^6 - 2x = 3y^6 + y^5 - y^2$, there is no like shooting fish in a barrel way to solve for $y$ explicitly in terms of $x$. Still, at that place may exist functions $f(10)$ so that if we plug $y=f(x)$ into this equation, information technology holds true. The possible functions $f(x)$ that take this property are and so said to be defined implicitly by the given equation.

With the assumption that a given equation implicitly defines at least one differentiable function, ane can find the derivative of ane variable in terms of some other through a process known as implicit differentiation.

Consider the post-obit:

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Suppose we are asked to find $\displaystyle{\frac{dy}{dx}}$ given

$$3x^four y^ii - 7xy^iii = iv - 8y$$

Our strategy is to treat the left and right sides of the equation above every bit functions of $x$ in their own right, and differentiate them both with respect to $ten$. Equally the left and right sides agree, their derivatives should be equal likewise. This will requite us an equation in terms of an unknown $dy/dx$ for which we may and then solve.

Importantly, every bit we differentiate each side, nosotros treat $y$ as a function of $x$. This requires that whenever we accept a derivative of a function of $y$ (e.grand., $y^2$, or $y^three$), nosotros must invoke the concatenation dominion and pick up an additional gene of $dy/dx$.

Thus, as an case, just as $\displaystyle{\frac{d}{dx}\left([f(ten)]^three\right) = 3[f(x)]^two \cdot f'(10)}$, nosotros discover $\displaystyle{\frac{d}{dx}\left(y^3\right) = 3y^ii\frac{dy}{dx}}$.

Differentiating in this fashion both sides of the equation to a higher place results in the following (after applying the product dominion twice):

$$\left[ 12x^3 y^2 + 3x^four \cdot 2y \frac{dy}{dx} \right] - \left[ 7y^3 + 7x \cdot 3y^ii \frac{dy}{dx} \correct] = 0 - 8 \frac{dy}{dx}$$

Note how the consequence is linear in terms of the $dy/dx$. Nicely, if the original equation is in terms of only $x$s and $y$s, this always happens! Equally such, we tin can always easily solve for this unknown $dy/dx$, by collecting all of the $dy/dx$ terms on one side, with all other terms on the other. Doing this here produces the following:

$$(6x^iv y - 21 xy^ii + eight) \cdot \frac{dy}{dx} = 7y^3 - 12x^3 y^ii$$

Dividing by the factor to the left of $dy/dx$, nosotros and so accept

$$\frac{dy}{dx} = \frac{7y^three - 12x^iii y^2}{6x^4y-21xy^2+8}$$

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As one tin see, the resulting expression for $dy/dx$ does involve $y$ values. This almost e'er happens -- but it should be no surprise, as the original equation we were given may actually implicitly define multiple functions.

To understand this better, suppose nosotros started instead with $x^2 + y^2 = 25$, whose graph is a circle. The ii functions $f_1$ and $f_2$ that are implicitly defined by this equation correspond to the semi-circles on the top and lesser halves of this circumvolve, equally shown below. For any given $x$ value (e.g., $x=three$), nosotros thus have two possible tangent lines with two potentially dissimilar slopes (and hence, values of $dy/dx$). To find the slope of a tangent line to the curve at some $x$-value, nosotros demand to know to which function we hope to be tangent. The $y$-value gives us this data.

3 Y 2 7xy 4,

Source: http://math.oxford.emory.edu/site/math111/implicitDifferentiation/

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