Welcome to Chapter 3-4 and 3-5!
In this session, we are expanding our derivative toolkit. We move beyond simple polynomials and standard exponentials to tackle more complex compositions of algebraic functions. Specifically, we explore how to handle general exponential bases and the powerful technique known as Implicit Differentiation.
1. Derivatives of Exponentials and General Bases
We already know that the derivative of $e^x$ is simply $e^x$. But what happens when the base is a number other than $e$, like $2^x$ or $5^{x^2}$? Or when we have a function inside the exponent?
We combine the Chain Rule with our new rules for exponentials:
- Base $e$: $\frac{d}{dx} e^u = e^u \cdot u'$
- General Base $b$: $\frac{d}{dx} b^u = (\ln b) \cdot b^u \cdot u'$
Example from class notes: For $h(x) = 5^{x^2}$, we identify the inner function $u = x^2$ (so $u' = 2x$). Using the general base rule, the derivative is $h'(x) = (\ln 5) \cdot 5^{x^2} \cdot (2x)$.
2. The Square Root Shortcut
A recurring theme in our notes is the "Square Root Shortcut." Instead of converting a radical to the $\frac{1}{2}$ power every time, you can use this efficient formula for the Chain Rule:
$$ \frac{d}{dx} \sqrt{u} = \frac{u'}{2\sqrt{u}} $$
This is incredibly useful for algebraic simplification, as seen in our class example $y = \sqrt{3x^k - 4x}$.
3. Implicit Differentiation
Up until now, we have dealt with explicit functions where $y$ is isolated on one side (e.g., $y = x^2 + 5$). But what about equations like $x^2 + y^2 = 25$ or complex relations like $3x^2 + 3y^2y' = 6y + 6xy'$?
When $y$ is mixed in with $x$, we use Implicit Differentiation. The key concept to remember is that $y$ is a function of $x$. Therefore, whenever you differentiate a $y$ term, you must apply the Chain Rule and multiply by $y'$ (or $\frac{dy}{dx}$).
Step-by-Step Strategy:
- Differentiate both sides of the equation with respect to $x$.
- Remember to add $y'$ whenever you take the derivative of a term containing $y$.
- Watch out for the Product Rule! If you have a term like $6xy$, you are multiplying two functions. The derivative is $6(1 \cdot y + x \cdot y')$.
- Collect all terms with $y'$ on one side of the equation and non-$y'$ terms on the other.
- Factor out $y'$ and divide to solve for the final derivative.
Key Takeaway
Implicit differentiation allows us to find the slope of tangent lines for curves that aren't simple functions, such as circles or ellipses. It requires strong algebra skills, particularly when grouping terms, but it unlocks the ability to solve a much wider range of geometric problems.
Keep practicing those Product Rule applications within implicit problems—that is the most common place to make a small sign error! Happy calculating!