Solution:
Let \(u = f(x)\), then \(y = u^3\)
Using the chain rule:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]We have:
\[\frac{dy}{du} = 3u^2 \text{ and } \frac{du}{dx} = f'(x)\]Therefore:
\[y' = 3u^2 \cdot f'(x) = 3(f(x))^2 \cdot f'(x)\]Definition: An explicit function is one where the dependent variable \(y\) is written directly in terms of the independent variable \(x\), in the form \(y = f(x)\).
Examples of explicit functions:
With explicit functions, we can compute derivatives using the differentiation rules we've learned (power rule, chain rule, product rule, quotient rule, etc.).
Definition: An implicit function is one where \(y\) is still dependent on \(x\), but you do not have a formula directly given with \(y\) explicitly written in terms of only \(x\).
Examples of implicit functions:
Important: In many cases, it is very difficult or impractical to solve for \(y\) explicitly. However, we can still compute derivatives using implicit differentiation!
Solution:
Step 1: Take the derivative with respect to \(x\) on both sides:
\[\frac{d}{dx}(x^2 + y^3) = \frac{d}{dx}(5)\]Applying the derivative:
\[2x + \frac{d}{dx}(y^3) = 0\]For the term \(\frac{d}{dx}(y^3)\), we use the chain rule. Think of it as \(\frac{d}{dx}((y(x))^3)\):
\[\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx} = 3y^2 y'\]So our equation becomes:
\[2x + 3y^2 y' = 0\]Step 2: Solve for \(y'\):
\[3y^2 y' = -2x\] \[y' = \frac{-2x}{3y^2}\]Verification: We can verify this result by solving explicitly. From \(x^2 + y^3 = 5\), we get \(y = (5 - x^2)^{1/3}\).
Using the chain rule:
\[y' = \frac{d}{dx}(5 - x^2)^{1/3} = \frac{1}{3}(5 - x^2)^{-2/3} \cdot (-2x) = \frac{-2x}{3(5 - x^2)^{2/3}}\]Since \(y^3 = 5 - x^2\), we have \(y = (5 - x^2)^{1/3}\), so \(y^2 = (5 - x^2)^{2/3}\).
Therefore: \(y' = \frac{-2x}{3y^2}\) ✓
Solution:
From Example 1, we know that:
\[y' = \frac{-2x}{3y^2}\]At the point \((2, 1)\), the slope is:
\[y'\bigg|_{(2,1)} = \frac{-2(2)}{3(1)^2} = \frac{-4}{3}\]Using the point-slope form of a line:
\[y - 1 = -\frac{4}{3}(x - 2)\] \[y - 1 = -\frac{4}{3}x + \frac{8}{3}\] \[y = -\frac{4}{3}x + \frac{11}{3}\]Graph Description: The curve \(x^2 + y^3 = 5\) passes through the point \((2, 1)\). The tangent line at this point has slope \(-4/3\) and touches the curve at exactly one point locally.
Solution:
Step 1: Take the derivative with respect to \(x\) on both sides:
\[\frac{d}{dx}(e^y) = \frac{d}{dx}(x\sin y)\]For the left side, using the chain rule:
\[\frac{d}{dx}(e^y) = e^y \cdot y'\]For the right side, using the product rule:
\[\frac{d}{dx}(x\sin y) = \frac{dx}{dx} \cdot \sin y + x \cdot \frac{d}{dx}(\sin y)\] \[= \sin y + x \cdot \cos y \cdot y'\]So we have:
\[e^y \cdot y' = \sin y + x\cos y \cdot y'\]Step 2: Solve for \(y'\):
\[e^y \cdot y' - x\cos y \cdot y' = \sin y\] \[y'(e^y - x\cos y) = \sin y\] \[y' = \frac{\sin y}{e^y - x\cos y}\]Solution:
Step 1: Take the derivative with respect to \(x\) on both sides:
\[\frac{d}{dx}[\sin(xy)] = \frac{d}{dx}(x + y)\]For the left side, using the chain rule:
\[\frac{d}{dx}[\sin(xy)] = \cos(xy) \cdot \frac{d}{dx}(xy)\]Using the product rule on \(xy\):
\[\frac{d}{dx}(xy) = y + xy'\]So:
\[\cos(xy) \cdot (y + xy') = 1 + y'\]Expanding:
\[y\cos(xy) + x\cos(xy) \cdot y' = 1 + y'\]Step 2: Solve for \(y'\):
\[x\cos(xy) \cdot y' - y' = 1 - y\cos(xy)\] \[y'[x\cos(xy) - 1] = 1 - y\cos(xy)\] \[y' = \frac{1 - y\cos(xy)}{x\cos(xy) - 1}\]Alternatively, by factoring out \(-1\):
\[y' = \frac{y\cos(xy) - 1}{1 - x\cos(xy)}\]Solution:
First, find \(y'\):
Step 1: Take the derivative with respect to \(x\):
\[\frac{d}{dx}(x^2 + xy - y^3) = \frac{d}{dx}(7)\] \[2x + \frac{d}{dx}(xy) - \frac{d}{dx}(y^3) = 0\]Using the product rule on \(xy\) and chain rule on \(y^3\):
\[2x + (y + xy') - 3y^2y' = 0\] \[2x + y + xy' - 3y^2y' = 0\]Step 2: Solve for \(y'\):
\[xy' - 3y^2y' = -2x - y\] \[y'(x - 3y^2) = -(2x + y)\] \[y' = \frac{-(2x + y)}{x - 3y^2} = \frac{-2x - y}{x - 3y^2}\]Now find \(y'' by differentiating the equation \(2x + y + xy' - 3y^2y' = 0\) again:
\[\frac{d}{dx}(2x + y + xy' - 3y^2y') = 0\]Term by term:
\[2 + y' + \frac{d}{dx}(xy') - \frac{d}{dx}(3y^2y') = 0\]For \(\frac{d}{dx}(xy')\), use the product rule:
\[\frac{d}{dx}(xy') = y' + xy''\]For \(\frac{d}{dx}(3y^2y')\), use the product rule with chain rule:
\[\frac{d}{dx}(3y^2y') = 3\left(\frac{d}{dx}(y^2) \cdot y' + y^2 \cdot y''\right)\] \[= 3(2yy' \cdot y' + y^2y'') = 3(2y(y')^2 + y^2y'') = 6y(y')^2 + 3y^2y''\]Putting it together:
\[2 + y' + y' + xy'' - 6y(y')^2 - 3y^2y'' = 0\] \[2 + 2y' - 6y(y')^2 + xy'' - 3y^2y'' = 0\] \[2 + 2y' - 6y(y')^2 + y''(x - 3y^2) = 0\]Solving for \(y''\):
\[y''(x - 3y^2) = -(2 + 2y' - 6y(y')^2)\] \[y'' = \frac{-(2 + 2y' - 6y(y')^2)}{x - 3y^2}\]We can substitute \(y' = \frac{-2x - y}{x - 3y^2}\) into this expression if we want \(y''\) entirely in terms of \(x\) and \(y\).
Key Takeaway: Implicit differentiation allows us to find derivatives (and even higher-order derivatives) without explicitly solving for \(y\) in terms of \(x\). The result is typically expressed in terms of both \(x\) and \(y\).