7.5 Implicit differentiation
When \(y\) is not given explicitly in terms of \(x\), differentiate both sides with respect to \(x\) and treat \(y\) as a function of \(x\). Every time a \(y\) is differentiated, the chain rule contributes a factor \(\frac {dy}{dx}\).
Solution. Differentiate each term with respect to \(x\). \begin {align*} \frac {d}{dx}\left (x^2\right ) &= 2x\\ \frac {d}{dx}(4xy) &= 4y+4x\frac {dy}{dx} &&\text {(product rule)}\\ \frac {d}{dx}\left (3y^2\right ) &= 6y\frac {dy}{dx} &&\text {(chain rule)}\\ \frac {d}{dx}(10) &= 0 \end {align*}
Putting these together: \[2x+4y+4x\frac {dy}{dx}-6y\frac {dy}{dx}=0.\]
Now collect the \(\frac {dy}{dx}\) terms and factor: \[\frac {dy}{dx}\left (4x-6y\right )=-2x-4y\] \[\implies \quad \frac {dy}{dx}=\frac {-(2x+4y)}{4x-6y}=\frac {-(x+2y)}{2x-3y}.\]
Note 7.22. The answer contains both \(x\) and \(y\), and that is normal for implicit differentiation — it is not an unfinished answer. The two places to be careful are the \(4xy\) term, which needs the product rule because both factors involve the variables, and remembering the \(\frac {dy}{dx}\) every time a \(y\) is differentiated.
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