So far, we've built a strong toolkit for finding derivatives of polynomials, exponential functions, and trigonometric functions. Today, we add the final key pieces to our collection: logarithmic and inverse trigonometric functions.
Along the way, we'll discover a powerful new technique called logarithmic differentiation. This method can transform a messy problem involving products, quotients, and powers into a much simpler one. It also allows us to tackle a new type of function we couldn't differentiate before.
Our first goal is to find the derivative of the natural logarithmic function, $f(x) = \ln(x)$. We can find this by cleverly using its inverse relationship with the exponential function, $e^x$, and applying implicit differentiation.
Let $y = \ln(x)$. To solve for $x$, we can exponentiate both sides using base $e$:
Recall that the exponential function $g(x)=e^x$ and the natural log function $f(x)=\ln(x)$ are inverses. By definition, this means that $g(f(x)) = e^{\ln(x)} = x$. Applying this property, our equation simplifies to:
Now, we can differentiate this implicit relation with respect to $x$:
Remembering the chain rule for the left side, we get:
Now, we solve for $\frac{dy}{dx}$:
This is the derivative, but it's in terms of $y$. We want it in terms of $x$. Since we established that $e^y = x$, we can substitute $x$ back into the expression:
This gives us our fundamental rule for the derivative of the natural logarithm.
The graph below shows the function $f(x) = \ln(x)$ and its tangent line at a movable point $(a, \ln a)$. Notice how the slope of the tangent line, $m$, is always equal to the value of $1/a$. This provides a strong visual confirmation that $\frac{d}{dx}(\ln x) = \frac{1}{x}$.
When we combine this with the chain rule, we get the more general and useful form. If $u$ is a differentiable function of $x$:
To find the derivative for a logarithm with a different base, like $\log_a(x)$, we can use the change of base formula: $\log_a(x) = \frac{\ln(x)}{\ln(a)}$. Since $\ln(a)$ is just a constant, the derivative is:
Find the derivative of $f(x) = \ln(x^3 + \cos x)$.
Logarithmic differentiation is not a formula to memorize, but rather a powerful method to use in two specific situations where normal differentiation rules are either extremely cumbersome or impossible to apply.
When faced with a function made of many pieces being multiplied, divided, or raised to powers, using the Product, Quotient, and Chain rules can be a nightmare. Logarithmic differentiation simplifies this by using properties of logarithms to break the function apart first.
Differentiate $y = \frac{\sqrt{x+1} \cdot (x-2)^3}{(x+5)^4}$.
Step 1: Take the natural logarithm of both sides.
Before moving to Step 2, let's recall the key properties of logarithms. The quotient rule states $\ln(a/b) = \ln(a) - \ln(b)$, and the product rule states $\ln(a \cdot b) = \ln(a) + \ln(b)$. Combining these, an expression like $\ln(\frac{a \cdot b}{c})$ becomes $\ln(a \cdot b) - \ln(c)$, which then expands to $\ln(a) + \ln(b) - \ln(c)$. We will apply this logic to our function.
Step 2: Use log properties to expand the right side.
Step 3: Differentiate both sides implicitly with respect to $x$.
Step 4: Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$.
Finally, substitute the original expression for $y$ back in.
This is a critical case. When both the base and the exponent are variable functions, neither the Power Rule nor the Exponential Rule applies. For example, for $y=x^{\sin x}$, we can't use the power rule (that's for $x^n$) or the exponential rule (that's for $a^x$). Logarithmic differentiation is the essential tool here.
Differentiate $y = x^{\sin x}$.
Step 1: Take the natural log of both sides.
Step 2: Use the log property to bring the exponent down.
Step 3: Differentiate both sides implicitly. The right side now requires the Product Rule.
Step 4: Solve for $\frac{dy}{dx}$.
Find the derivative of $y = (\sin x)^{\ln x}$.
Lastly, we will find the derivatives for the inverse trigonometric functions. The process is similar to how we found the derivative of $\ln x$: we use implicit differentiation.
Let $y = \arcsin(x)$. To have an inverse, a function must be one-to-one. Since $f(x)=\sin(x)$ is not one-to-one over its full domain, we restrict its domain to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to define its inverse, the arcsin function. So, by definition, $y = \arcsin(x)$ is the value $y$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(y) = x$.
We can see this by taking the sine of both sides of our starting equation:
Because the functions are defined as inverses, this simplifies to:
Differentiating this implicitly with respect to $x$ gives:
To express $\cos(y)$ in terms of $x$, we can use the identity $\sin^2(y) + \cos^2(y) = 1$, which gives $\cos(y) = \sqrt{1 - \sin^2(y)}$. Because of the range restriction on $y$, $\cos(y)$ is non-negative, so we take the positive root. Since $\sin(y) = x$, we have:
Substituting this back gives our final rule:
Similarly, for $y = \arctan(x)$, we have $\tan(y) = x$. Differentiating gives:
Using the identity $\sec^2(y) = 1 + \tan^2(y)$, and knowing $\tan(y) = x$, we get:
This graph shows the function $f(x) = \arctan(x)$ and its tangent line. As you move the point, observe the slope $m$. The slope is steepest at $x=0$ (where $m=1$) and flattens out as $x$ moves towards $\pm\infty$. This perfectly matches our derivative formula, $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$, which is maximized at $x=0$.
The derivatives for all six inverse trig functions can be found this way. Notice the helpful pattern: the derivative of a "co-" function is the negative of its counterpart.
| Function | Derivative |
|---|---|
| $f(x) = \arcsin(x)$ | $f'(x) = \frac{1}{\sqrt{1-x^2}}$ |
| $f(x) = \arccos(x)$ | $f'(x) = -\frac{1}{\sqrt{1-x^2}}$ |
| $f(x) = \arctan(x)$ | $f'(x) = \frac{1}{1+x^2}$ |
| $f(x) = \operatorname{arccot}(x)$ | $f'(x) = -\frac{1}{1+x^2}$ |
| $f(x) = \operatorname{arcsec}(x)$ | $f'(x) = \frac{1}{x\sqrt{x^2-1}}$ |
| $f(x) = \operatorname{arccsc}(x)$ | $f'(x) = -\frac{1}{x\sqrt{x^2-1}}$ |
Find the derivative of $y = \arcsin(e^{2x})$.
Here, the inside function is $u = e^{2x}$, and its derivative is $u' = 2e^{2x}$.
Using the formula for the derivative of $\arcsin u$, we get:
Find the derivative of $f(x) = x^2 \arctan(3x)$.
Find the derivative of $y = \ln(x^2 - e^x)$.
Find the derivative of $y = (\cos x)^{x^2}$.
Find the derivative of $g(t) = t \cdot \operatorname{arcsec}(t^2)$.
Find the derivative of $y = x^x$.
In this section, we completed our set of basic differentiation rules by deriving the formulas for logarithmic and inverse trigonometric functions. We showed that $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and derived all six inverse trig derivatives using implicit differentiation. More importantly, we introduced the powerful logarithmic differentiation technique, which simplifies complex product/quotient problems and is essential for differentiating functions of the form $[f(x)]^{g(x)}$.
After this lecture, you should be able to: