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3.10 Derivatives of Inverse Functions

(NOT on Exam 2)

\[ y = \sin x \rightarrow y' = \cos x \]

\[ y = \sin^{-1} x \rightarrow y' = ? \]

We can accomplish this by doing implicit differentiation:

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

Differentiate implicitly:

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

Note: \( \sin(y) \) is a function of \( x \), so we need the Chain Rule: \( \frac{dy}{dx} \)

\[ \cos(y) \frac{dy}{dx} = 1 \]

\[ \frac{dy}{dx} = \frac{1}{\cos(y)} \]

Usually, if possible, we want the result explicitly (as a function of \( x \), not \( y \)).

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We don't want that \( \cos(y) \) on the right.

We know: \[ \cos^2(y) + \sin^2(y) = 1 \]

From earlier, \( \sin(y) = x \), so \( \sin^2(y) = x^2 \)

\[ \cos^2(y) + x^2 = 1 \]
\[ \text{or } \cos^2(y) = 1 - x^2 \quad \text{or } \cos(y) = \sqrt{1 - x^2} \]

Sub into:

\[ \frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1 - x^2}} \]

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

Chain rule extension

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

Following similar steps, we get:

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

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

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Example: Derivative of an Inverse Sine Function

Find the derivative of:

\[ y = \sin^{-1}(6x^5) \]

There are two options to solve this: use the derivative formula for \(\sin^{-1}(u)\) or use implicit differentiation.

Option 1: Using the Formula

The formula for the derivative of the inverse sine function is:

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

Applying this to our function where \( u = 6x^5 \):

\[ y' = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} = \frac{1}{\sqrt{1-(6x^5)^2}} \frac{d}{dx}(6x^5) \]

Final Result (Option 1):

\[ y' = \frac{30x^4}{\sqrt{1-36x^{10}}} \]

Option 2: Implicit Differentiation

Start with the original equation and take the sine of both sides:

\[ y = \sin^{-1}(6x^5) \]\[ \sin(y) = 6x^5 \]

Now, differentiate both sides with respect to \( x \):

\[ \frac{d}{dx} \sin(y) = \frac{d}{dx}(6x^5) \]\[ \cos(y) y' = 30x^4 \]

Solve for \( y' \):

\[ y' = \frac{30x^4}{\cos(y)} \]
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Eliminating \(\cos(y)\)

Now we need to eliminate \(\cos(y)\) from our expression for \(y'\). We use the Pythagorean identity:

\[ \cos^2(y) + \sin^2(y) = 1 \]

From our earlier work in the implicit differentiation step, we know that:

\[ \sin(y) = 6x^5 \]\[ \text{so, } \sin^2(y) = (6x^5)^2 = 36x^{10} \]

Substituting this back into the identity:

\[ \cos^2(y) = 1 - 36x^{10} \]\[ \cos(y) = \sqrt{1 - 36x^{10}} \]

Then, substituting this back into our expression for \( y' \):

\[ y' = \frac{30x^4}{\cos(y)} \]

Final Result (Option 2):

\[ y' = \frac{30x^4}{\sqrt{1-36x^{10}}} \]
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Example: Derivative of an Inverse Trigonometric Function

Consider the function:

\[ y = \cos^{-1}(\sin(x^2)) \]

Using the chain rule where \( u = \sin(x^2) \), we recall the derivative formula for inverse cosine:

\[ \frac{d}{dx} \cos^{-1}(u) = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx} \]

Applying this to our function:

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

For the second part, we use the chain rule again where \( u = x^2 \). Recall that:

\[ \frac{d}{dx} \sin(u) = \cos(u) \frac{du}{dx} \]

Continuing the calculation:

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

Simplification

We can simplify it a bit further using the Pythagorean identity:

\[ \sin^2(x^2) + \cos^2(x^2) = 1 \]\[ \cos^2(x^2) = 1 - \sin^2(x^2) \]

Substituting this back into our derivative:

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

Need absolute value because the sign of \( \cos(x^2) \) is not known.

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Relationship Between Tangent Slopes of a Function and Its Inverse

What is the relationship between the tangent line slope on a function and its inverse?

Consider a function \( y = f(x) \) and its inverse \( y = f^{-1}(x) \). The graph shows these functions reflected across the line \( y = x \). A point \( (x_0, y_0) \) on the original function corresponds to the point \( (y_0, x_0) \) on the inverse function.

Coordinate graph showing y=f(x), y=f-inverse(x), and the line y=x with points (x0, y0) and (y0, x0).

When \( f'(x_0) \) is steep, the corresponding tangent line slope on \( f^{-1}(x) \) is shallow, and when \( f'(x_0) \) is shallow, the slope on \( f^{-1}(x) \) is steep.

This suggests the relationship is a reciprocal relationship.

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What does it look like?

Starting with the relationship:

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

Differentiate implicitly with respect to \(x\):

\[ \frac{d}{dx}(x) = \frac{d}{dx} f^{-1}(y) \]\[ 1 = (f^{-1})'(y) \cdot \frac{dy}{dx} \]

From chain rule, since \(y = f(x)\)

Since \(\frac{dy}{dx} = f'(x)\), we can sub it in and solve for \((f^{-1})'(y)\):

\[ 1 = (f^{-1})'(y) \cdot f'(x) \]

So, we get:

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

Notice \(f'\) depends on \(x\) and \((f^{-1})'\) depends on the corresponding \(y\).

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Example

Given \(f(x) = e^x\), without finding \(f^{-1}(x)\), find \((f^{-1})'(e)\).

Formula:

\[ (f^{-1})'(y) = \frac{1}{f'(x)} \]
\[ (f^{-1})'(e) = \frac{1}{f'(x)} \]

\(y = e\)

corresponding \(x\) on \(f(x) = e^x\) such that \(y = e\)

\(f(x) = e^x\) and \(f = y = e\)

Then \(e = e^x \rightarrow x = 1\)

\(f(x) = e^x\)

\(f'(x) = e^x\)

\(f'(1) = e\)

Then,

\[ (f^{-1})'(e) = \frac{1}{f'(1)} = \frac{1}{e} \]
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Let's verify that by finding the inverse and then its derivative at \(e\)

\[f(x) = e^x \rightarrow f^{-1}(x) = \ln(x)\]
\[(f^{-1})'(x) = \frac{1}{x}\]
\[\text{so, } (f^{-1})'(e) = \frac{1}{e}\]

same as using the other method