Let’s solve problem 40:
Given: \( 3y = x e^{5y} \)
Find: \( \frac{dy}{dx} \), assuming \( y \) is a differentiable function of \( x \).
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We’ll use
implicit differentiation, because \( y \) is not isolated — it’s mixed in with \( x \) on both sides.
Step 1: Differentiate both sides with respect to \( x \)
Left side:
\( \frac{d}{dx}(3y) = 3 \cdot \frac{dy}{dx} \) → (because derivative of \( y \) w.r.t. \( x \) is \( \frac{dy}{dx} \))
Right side:
\( \frac{d}{dx}(x e^{5y}) \) → This is a product of two functions: \( u = x \), \( v = e^{5y} \)
Use the
product rule: \( (uv)' = u'v + uv' \)
- \( u = x \) → \( u' = 1 \)
- \( v = e^{5y} \) → To differentiate this, we need the
chain rule because \( y \) is a function of \( x \)
Derivative of \( e^{5y} \) w.r.t. \( x \):
= \( e^{5y} \cdot \frac{d}{dx}(5y) = e^{5y} \cdot 5 \cdot \frac{dy}{dx} \)
So now apply product rule:
\( \frac{d}{dx}(x e^{5y}) = (1)(e^{5y}) + (x)(5 e^{5y} \cdot \frac{dy}{dx}) \)
Simplify:
= \( e^{5y} + 5x e^{5y} \cdot \frac{dy}{dx} \)
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Step 2: Set left and right derivatives equal
From earlier:
Left: \( 3 \frac{dy}{dx} \)
Right: \( e^{5y} + 5x e^{5y} \frac{dy}{dx} \)
So:
\[
3 \frac{dy}{dx} = e^{5y} + 5x e^{5y} \frac{dy}{dx}
\]
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Step 3: Solve for \( \frac{dy}{dx} \)
Get all terms with \( \frac{dy}{dx} \) on one side.
Subtract \( 5x e^{5y} \frac{dy}{dx} \) from both sides:
\[
3 \frac{dy}{dx} - 5x e^{5y} \frac{dy}{dx} = e^{5y}
\]
Factor out \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} \left( 3 - 5x e^{5y} \right) = e^{5y}
\]
Now divide both sides by \( (3 - 5x e^{5y}) \):
\[
\frac{dy}{dx} = \frac{e^{5y}}{3 - 5x e^{5y}}
\]
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This is the derivative. We don’t simplify further unless asked — and the worksheet says “answers are not simplified,” so this is fine.
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Final Answer:
\[
\boxed{\frac{e^{5y}}{3 - 5x e^{5y}}}
\]
Parent Tip: Review the logic above to help your child master the concept of ap calculus derivative worksheet.
