Note: This material is NOT on Exam 2
Topics covered:
If \(\sin y = x\), find \(\cos y\) and \(\tan y\).
Solution:
Using a right triangle where:
Right Triangle: A right triangle with hypotenuse 1, opposite side \(x\), and adjacent side \(\sqrt{1-x^2}\). The angle \(y\) is at the base of the triangle.
Therefore:
\[\cos y = \frac{\text{adj}}{\text{hyp}} = \frac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2}\] \[\tan y = \frac{\text{opp}}{\text{adj}} = \frac{x}{\sqrt{1-x^2}}\]If \(\tan y = x\), find \(\sin y\) and \(\cos y\).
Solution:
Using a right triangle where:
Right Triangle: A right triangle with adjacent side 1, opposite side \(x\), and hypotenuse \(\sqrt{1+x^2}\). The angle \(y\) is at the base of the triangle.
Therefore:
\[\sin y = \frac{\text{opp}}{\text{hyp}} = \frac{x}{\sqrt{1+x^2}}\] \[\cos y = \frac{\text{adj}}{\text{hyp}} = \frac{1}{\sqrt{1+x^2}}\] \[\sec y = \frac{1}{\cos y} = \sqrt{1+x^2}\]WARNING: Be careful with notation!
If \(\sin y = x\), then \(y = \sin^{-1}(x)\)
This does NOT mean: \(\sin^{-1}(x) = \frac{1}{\sin(x)}\)
The notation \(\sin^{-1}(x)\) means the inverse function of sine, not the reciprocal.
If we want the derivative \(\frac{d}{dx}[\sin^{-1}(x)]\), we will find it using the relationship between a function and its inverse.
Reference Triangle: Right triangle with hypotenuse 1, opposite side \(x\), adjacent side \(\sqrt{1-x^2}\), showing that \(\cos y = \sqrt{1-x^2}\).
Formula:
\[\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}\]Reference Triangle: Right triangle with hypotenuse 1, adjacent side \(x\), opposite side \(\sqrt{1-x^2}\), showing that \(\sin y = \sqrt{1-x^2}\).
Formula:
\[\frac{d}{dx}[\cos^{-1}(x)] = \frac{-1}{\sqrt{1-x^2}}\]Reference Triangle: Right triangle with adjacent side 1, opposite side \(x\), and hypotenuse \(\sqrt{1+x^2}\), showing that \(\sec y = \sqrt{1+x^2}\) and \(\cos y = \frac{1}{\sqrt{1+x^2}}\).
Formula:
\[\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}\]Summary: Derivatives of Inverse Trigonometric Functions
\[\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}\] \[\frac{d}{dx}[\cos^{-1}(x)] = \frac{-1}{\sqrt{1-x^2}}\] \[\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}\]Find \(y'\) if \(y = \cos^{-1}(8x^2)\).
Solution:
Using the chain rule, we need to find the derivative of the outer function \(\cos^{-1}(u)\) where \(u = 8x^2\), then multiply by the derivative of the inner function.
Step 1: Identify the composition
Let \(u = 8x^2\), so \(y = \cos^{-1}(u)\)
Step 2: Apply chain rule
\[\frac{dy}{dx} = \frac{d}{du}[\cos^{-1}(u)] \cdot \frac{du}{dx}\]Step 3: Find each derivative
\[\frac{d}{du}[\cos^{-1}(u)] = \frac{-1}{\sqrt{1-u^2}}\] \[\frac{du}{dx} = \frac{d}{dx}[8x^2] = 16x\]Step 4: Substitute back
\[y' = \frac{-1}{\sqrt{1-(8x^2)^2}} \cdot 16x = \frac{-16x}{\sqrt{1-64x^4}}\]Reference Triangle for \(\cos^{-1}(8x^2)\): A right triangle with hypotenuse 1, adjacent side \(8x^2\), and opposite side \(\sqrt{1-(8x^2)^2}\), which shows that \(\sin y = \sqrt{1-(8x^2)^2}\).
Final Answer:
\[y' = \frac{-16x}{\sqrt{1-64x^4}}\]There is a general relationship between the derivative of a function and the derivative of its inverse function.
Theorem: If \(y = f(x)\) and \(f^{-1}(y) = x\), then:
\[\frac{d}{dx}[f^{-1}(x)] = \frac{1}{\frac{dy}{dx}}\text{ evaluated at the corresponding point}\]Or equivalently, if \(y = f^{-1}(x)\), then \(f(y) = x\), and taking derivatives:
\[\frac{d}{dx}[f(y)] = \frac{d}{dx}[x]\] \[f'(y) \cdot \frac{dy}{dx} = 1\] \[\frac{dy}{dx} = \frac{1}{f'(y)}\]This relationship was used to derive the formulas for inverse trigonometric functions:
General Formula: If \(y = f(x)\) and \(f(y) = x\), then:
\[\boxed{\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(y)} = \frac{1}{\frac{dy}{dx}\text{ of } f}}\]The graphs of a function \(y = f(x)\) and its inverse \(y = f^{-1}(x)\) are mirror images of each other across the line \(y = x\).
Graph Description: The graph shows three curves:
The two curves are reflections of each other across the line \(y = x\). If \((a, b = f(a))\) is a point on the curve of \(f\), then \((b, a = f^{-1}(b))\) is a point on the curve of \(f^{-1}\).
Key Insight: The mirror image relationship means that if the graph of \(f(x)\) has a certain slope at point \((a, b)\), then the graph of \(f^{-1}(x)\) will have the reciprocal slope at the mirrored point \((b, a)\).
Let \(f(x) = 9x + 5\), so \(f'(x) = 9\) and \(f(1) = 14\).
Find \((f^{-1})'(14)\).
Solution:
We know that \(f(1) = 14\), which means the point \((1, 14)\) is on the graph of \(f(x)\).
Therefore, the point \((14, 1)\) is on the graph of \(f^{-1}(x)\).
Using the relationship between derivatives:
\[(f^{-1})'(14) = \frac{1}{f'(1)}\]Since \(f'(1) = 9\):
\[(f^{-1})'(14) = \frac{1}{9}\]Interpretation:
Let \(f(x) = 8x^3 - 15x^2 - 5\). Find the derivative of the inverse function at the point \((-1, 2)\).
Solution:
Step 1: Understand what \((-1, 2)\) represents
The point \((-1, 2)\) is on the graph of the inverse \(f^{-1}\). This means:
Step 2: Verify that \(f(2) = -1\)
\[f(2) = 8(2)^3 - 15(2)^2 - 5 = 8(8) - 15(4) - 5 = 64 - 60 - 5 = -1\]Confirmed! ✓
Step 3: Find \(f'(x)\)
\[f'(x) = 24x^2 - 30x\]Step 4: Calculate \(f'(2)\)
\[f'(2) = 24(2)^2 - 30(2) = 24(4) - 60 = 96 - 60 = 36\]Step 5: Use the inverse function derivative formula
\[(f^{-1})'(-1) = \frac{1}{f'(2)} = \frac{1}{36}\]Final Answer:
\[(f^{-1})'(-1) = \frac{1}{36}\]Let \(f(x) = x^3 + x^2 + x + 1\). Find \(f^{-1}(4)\) and \((f^{-1})'(4)\).
Solution:
Part 1: Find \(f^{-1}(4)\)
We need to find \(x\) such that \(f(x) = 4\).
\[x^3 + x^2 + x + 1 = 4\]By guessing and checking, let's try \(x = 1\):
\[f(1) = 1^3 + 1^2 + 1 + 1 = 1 + 1 + 1 + 1 = 4\]Verified! ✓ Therefore:
\[f^{-1}(4) = 1\]Part 2: Find \((f^{-1})'(4)\)
First, find \(f'(x)\):
\[f'(x) = 3x^2 + 2x + 1\]Evaluate at \(x = 1\):
\[f'(1) = 3(1)^2 + 2(1) + 1 = 3 + 2 + 1 = 6\]Therefore, since \(f(1) = 4\) and \(f^{-1}(4) = 1\):
\[(f^{-1})'(4) = \frac{1}{f'(1)} = \frac{1}{6}\]Final Answer:
\[f^{-1}(4) = 1 \quad \text{and} \quad (f^{-1})'(4) = \frac{1}{6}\]Main Formulas:
Problem-Solving Strategy:
To find \((f^{-1})'(b)\):