Lesson 20 (3.10)

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Lesson 20 (3.10)

Today:

  • Derivatives of Inverse trig functions
  • Relation b/w derivative of function and its inverse

Office Hours: MWF: 130pm – 230pm, MATH 842

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  • Exam 2: Monday 10/19 8pm – 9pm (Instructions on Brightspace)
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Warmup:

\(\sin y = x\), find \(\cos y\), \(\tan y\).

Right triangle diagram for sin y = x. The bottom-left acute angle is labeled y. The vertical side opposite to angle y is labeled opp = x, the hypotenuse is labeled hyp = 1, and the horizontal base adjacent to angle y is labeled adj = \sqrt{1-x^2}.
Visual Description: Right triangle diagram for sin y = x. The bottom-left acute angle is labeled y. The vertical side opposite to angle y is labeled opp = x, the hypotenuse is labeled hyp = 1, and the horizontal base adjacent to angle y is labeled adj = \sqrt{1-x^2}.

\[ \text{adj}^2 + \text{opp}^2 = \text{hyp}^2 \]

\[ \text{adj} = \sqrt{1-x^2} \]

\[ \cos y = \frac{\text{adj}}{\text{hyp}} = \sqrt{1-x^2} \]

\[ \tan y = \frac{\text{opp}}{\text{adj}} = \frac{x}{\sqrt{1-x^2}} \]

\(\tan y = x\), find \(\sin y\), \(\cos y\).

Right triangle diagram for tan y = x. The bottom-left acute angle is labeled y. The vertical side opposite to angle y is labeled opp = x, the horizontal base adjacent to angle y is labeled adj = 1, and the hypotenuse is labeled hyp = \sqrt{1+x^2}.
Visual Description: Right triangle diagram for tan y = x. The bottom-left acute angle is labeled y. The vertical side opposite to angle y is labeled opp = x, the horizontal base adjacent to angle y is labeled adj = 1, and the hypotenuse is labeled hyp = \sqrt{1+x^2}.

\[ \text{hyp} = \sqrt{\text{opp}^2 + \text{adj}^2} = \sqrt{1+x^2} \]

\[ \sin y = \frac{\text{opp}}{\text{hyp}} = \frac{x}{\sqrt{1+x^2}} \]

\[ \cos y = \frac{\text{adj}}{\text{hyp}} = \frac{1}{\sqrt{1+x^2}} \]

\[ \sec y = \frac{1}{\cos y} = \sqrt{1+x^2} \]

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Derivative of \( \sin^{-1}(x) \)

\[ \sin^{-1}(x) \neq \frac{1}{\sin x} \]

\[ y = \sin^{-1}(x) \implies \sin y = x \]

Find \( y' = \frac{d}{dx} y = \frac{d}{dx} \sin^{-1}(x) \) using implicit differentiation

\[ \frac{d}{dx} \sin y = \frac{d}{dx} x \]

Right triangle diagram for angle y: The angle y is positioned at the acute lower-left vertex. The hypotenuse is labeled with length 1, the vertical side opposite to angle y is labeled x, and the horizontal base adjacent to angle y is labeled with length \(\sqrt{1 - x^2}\) by the Pythagorean theorem.
Visual Description: Right triangle diagram for angle y: The angle y is positioned at the acute lower-left vertex. The hypotenuse is labeled with length 1, the vertical side opposite to angle y is labeled x, and the horizontal base adjacent to angle y is labeled with length \(\sqrt{1 - x^2}\) by the Pythagorean theorem.

\[ \cos y \cdot y' = 1 \]

\[ y' = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}} \]

\[ \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \]

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Derivative of \( \cos^{-1}(x) \)

\( y = \cos^{-1}(x) \implies \cos(y) = x \)

Right triangle illustrating the relationship cos(y) = x. The acute angle at the bottom-left vertex is labeled y. The adjacent side along the horizontal base has length x, the hypotenuse has length 1, and the opposite vertical side is labeled with the square root of 1 minus x squared, determined via the Pythagorean theorem.
Visual Description: Right triangle illustrating the relationship cos(y) = x. The acute angle at the bottom-left vertex is labeled y. The adjacent side along the horizontal base has length x, the hypotenuse has length 1, and the opposite vertical side is labeled with the square root of 1 minus x squared, determined via the Pythagorean theorem.

take \( \frac{d}{dx} \) on both sides

\[ \frac{d}{dx} \cos y = \frac{d}{dx} x \] \[ -\sin y \cdot y' = 1 \] \[ y' = \frac{-1}{\sin y} = \frac{-1}{\sqrt{1-x^2}} \] \[ \frac{d}{dx} \cos^{-1}(x) = \frac{-1}{\sqrt{1-x^2}} \]

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Derivative of \(\tan^{-1}(x)\)

\[ y = \tan^{-1}(x) \rightsquigarrow \tan y = x \]

A right-angled triangle illustrating \(\tan y = x\). The bottom-left acute angle is labeled \(y\). The adjacent side along the base has length \(1\), the vertical opposite side has length \(x\), and the hypotenuse is labeled \(\sqrt{1+x^2}\).
Visual Description: A right-angled triangle illustrating \(\tan y = x\). The bottom-left acute angle is labeled \(y\). The adjacent side along the base has length \(1\), the vertical opposite side has length \(x\), and the hypotenuse is labeled \(\sqrt{1+x^2}\).

\(\frac{d}{dx}\) on both sides

\[ \frac{d}{dx} \tan y = \frac{d}{dx} x \]

\[ \sec^2 y \cdot y' = 1 \]

\[ y' = \frac{1}{\sec^2 y} \]

using triangle \(\sec y = \sqrt{1+x^2} \Rightarrow y' = \frac{1}{1+x^2}\)

\[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \]

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Example: \( y = \tan^{-1}(e^{2x}) \), find \( y' \)

Chain Rule: \( u = e^{2x} \)

\[ y' = \frac{d}{dx} \tan^{-1}(u) \] \[ = \frac{d}{du} \tan^{-1}(u) \cdot \frac{du}{dx} \] \[ = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] \[ = \frac{1}{1 + (e^{2x})^2} \frac{d}{dx} e^{2x} \] \[ = \frac{1}{1 + e^{4x}} \cdot 2e^{2x} \]

Using inverse function definition:

\[ y = \tan^{-1}(e^{2x}) \implies \tan y = e^{2x} \]

Right triangle diagram defining trigonometric functions of angle y. The adjacent base side is labeled 1, the opposite vertical side is labeled e^{2x}, and the hypotenuse is labeled \sqrt{1 + (e^{2x})^2}. The acute angle between the hypotenuse and the base is labeled y.
Visual Description: Right triangle diagram defining trigonometric functions of angle y. The adjacent base side is labeled 1, the opposite vertical side is labeled e^{2x}, and the hypotenuse is labeled \sqrt{1 + (e^{2x})^2}. The acute angle between the hypotenuse and the base is labeled y.

\( \frac{d}{dx} \) on both sides

\[ \sec^2 y \cdot y' = 2 \cdot e^{2x} \] \[ y' = \frac{2e^{2x}}{\sec^2 y} \]

Using triangle \( \sec y = \sqrt{1 + e^{4x}} \)

\[ y' = \frac{2e^{2x}}{1 + e^{4x}} \]

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Example: \( y = \cos^{-1}(8x^3) \), find \( y' \)

Use Chain Rule

\[ y' = \frac{d}{dx} \cos^{-1}(8x^3) \]

Let \( u = 8x^3 \):

\[ = \frac{d}{du} \cos^{-1}(u) \cdot \frac{du}{dx} \]

\[ = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]

\[ = \frac{-1}{\sqrt{1-(8x^3)^2}} \cdot \frac{d}{dx}(8x^3) \]

\[ = \frac{-24x^2}{\sqrt{1-64x^6}} \]

Alternative Method

\[ y = \cos^{-1}(8x^3) \implies \cos y = 8x^3 \]

A right triangle setup for the angle y, where cos(y) = 8x^3 / 1. The acute angle at the bottom-left is labeled y. The adjacent horizontal side is labeled 8x^3, the hypotenuse is labeled 1, and the vertical opposite side is labeled 'opp.' Below the triangle, the opposite side is computed via the Pythagorean theorem as opp = sqrt(1 - (adj)^2) = sqrt(1 - 64x^6).
Visual Description: A right triangle setup for the angle y, where cos(y) = 8x^3 / 1. The acute angle at the bottom-left is labeled y. The adjacent horizontal side is labeled 8x^3, the hypotenuse is labeled 1, and the vertical opposite side is labeled 'opp.' Below the triangle, the opposite side is computed via the Pythagorean theorem as opp = sqrt(1 - (adj)^2) = sqrt(1 - 64x^6).

\[ \text{opp} = \sqrt{1 - (\text{adj})^2} = \sqrt{1 - 64x^6} \]

Take \( \frac{d}{dx} \) on both sides:

\[ \frac{d}{dx} \cos y = \frac{d}{dx} 8x^3 \]

\[ (-\sin y) \cdot y' = 24x^2 \]

\[ y' = \frac{-24x^2}{\sin y} \]

Using triangle \( \sin y = \sqrt{1 - 64x^6} \):

\[ y' = \frac{-24x^2}{\sqrt{1 - 64x^6}} \]

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Relation between Derivative of a function & its inverse

\[ y = \ln x \rightsquigarrow e^y = x \rightsquigarrow e^y \cdot y' = 1 \rightsquigarrow y' = \frac{1}{e^y} \] \[ y = f^{-1}(x) \rightsquigarrow f(y) = x \rightsquigarrow f'(y) \cdot y' = 1 \rightsquigarrow y' = \frac{1}{f'(y)} \] \[ y = \sin^{-1}(x) \rightsquigarrow \sin y = x \rightsquigarrow \left(\frac{d}{dy}\sin y\right) \cdot y' = 1 \rightsquigarrow y' = \frac{1}{\left(\frac{d}{dy}\sin y\right)} \] \[ y = \cos^{-1} x \rightsquigarrow \cos y = x \rightsquigarrow y' = \frac{1}{\left(\frac{d}{dy}\cos y\right)} \] \[ y = \tan^{-1} x \rightsquigarrow \tan y = x \rightsquigarrow y' = \frac{1}{\left(\frac{d}{dy}\tan y\right)} \] \[ y = f^{-1}(x) \rightsquigarrow y' = \frac{1}{f'(y)} \rightsquigarrow \frac{d}{dx} f^{-1}(x) = \frac{1}{\frac{d}{dy} f(y)} \]

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What does this mean geometrically?

Cartesian coordinate plane illustrating the geometric relationship between a function and its inverse. The dashed line y = x represents the axis of symmetry. The curve y = f(x) is plotted, along with its reflection f^{-1}(x). Points (a, b) on y = f(x) are connected via dashed perpendicular blue lines across y = x to their corresponding reflected points (b, a) on f^{-1}(x). Red tangent lines drawn at (a, b) on f(x) and (b, a) on f^{-1}(x) illustrate reciprocal tangent slopes.
Visual Description: Cartesian coordinate plane illustrating the geometric relationship between a function and its inverse. The dashed line y = x represents the axis of symmetry. The curve y = f(x) is plotted, along with its reflection f^{-1}(x). Points (a, b) on y = f(x) are connected via dashed perpendicular blue lines across y = x to their corresponding reflected points (b, a) on f^{-1}(x). Red tangent lines drawn at (a, b) on f(x) and (b, a) on f^{-1}(x) illustrate reciprocal tangent slopes.

If \((a, b)\) is on graph of \(f(x)\), \((b, a)\) is on the graph of \(f^{-1}(x) \Rightarrow f(a) = b\),
\(f^{-1}(b) = a\)

\[ (f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}, \]

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eg: \( f(x) = 9x + 5 \), find \( (f^{-1})'(14) \)

find \( a \) such that \( f(a) = 14 \)

\[ 9a + 5 = 14 \implies a = 1 \]

\[ \left. \begin{array}{l} (1, 14) \text{ is on } f(x) \\ (14, 1) \text{ is on } f^{-1}(x) \end{array} \right\} f(1) = 14, \quad f^{-1}(14) = 1 \]

\[ (f^{-1})'(14) = \frac{1}{f'(1)} \]

\[ f(x) = 9x + 5 \implies f'(x) = 9 \implies f'(1) = 9 \]

\[ (f^{-1})'(14) = \frac{1}{9} \]

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Eg: \( f(x) = 8x^3 - 15x^2 - 5 \)

Find derivative of inverse at \( (-1, 2) \)

Point on the graph of inverse:
\( f^{-1}(-1) = 2 \) or \( f(2) = -1 \)

Check!

\[ f(2) = 8(2)^3 - 15(2)^2 - 5 \]

\[ = 64 - 60 - 5 = -1 \]

\[ f(-1) = -8 - 15 - 5 = -28 \neq 2 \]

\[ (f^{-1})'(-1) = \frac{1}{f'(2)} \]

\[ f'(x) = 24x^2 - 30x \rightsquigarrow f'(2) = 24(4) - 30(2) = 36 \]

\[ (f^{-1})'(-1) = \frac{1}{36} \]