Differentiating Inverse Functions - AP Calculus AB
Card 1 of 30
Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.
Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.
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No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.
No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.
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What is the derivative of $y = \text{arctan}(x)$?
What is the derivative of $y = \text{arctan}(x)$?
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$\frac{1}{1+x^2}$. Standard derivative formula for inverse tangent function.
$\frac{1}{1+x^2}$. Standard derivative formula for inverse tangent function.
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Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.
Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.
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No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.
No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.
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Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.
Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.
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$2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.
$2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.
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Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.
Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.
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$3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.
$3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.
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Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.
Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.
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$-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.
$-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.
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State the formula for the derivative of an inverse function.
State the formula for the derivative of an inverse function.
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$(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.
$(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.
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What is the derivative of $y = \text{arccot}(x)$?
What is the derivative of $y = \text{arccot}(x)$?
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$-\frac{1}{1+x^2}$. Standard derivative formula for inverse cotangent function.
$-\frac{1}{1+x^2}$. Standard derivative formula for inverse cotangent function.
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Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.
Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.
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$\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.
$\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.
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Which inverse function has a derivative of $-\frac{1}{1+x^2}$?
Which inverse function has a derivative of $-\frac{1}{1+x^2}$?
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$\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.
$\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.
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Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.
Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.
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$\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.
$\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.
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Which inverse function has a derivative of $\frac{1}{1+x^2}$?
Which inverse function has a derivative of $\frac{1}{1+x^2}$?
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$\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.
$\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.
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What is the domain of $y = \text{arccos}(x)$?
What is the domain of $y = \text{arccos}(x)$?
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$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
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What is the domain of $y = \text{arcsin}(x)$?
What is the domain of $y = \text{arcsin}(x)$?
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$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
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Find the derivative of $y = \arccot(6x)$ at $x = 0$.
Find the derivative of $y = \arccot(6x)$ at $x = 0$.
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$-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.
$-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.
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Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.
Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.
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$3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.
$3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.
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Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.
Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.
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$-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.
$-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.
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Find the derivative of $y = \text{arccot}(6x)$ at $x = 0$.
Find the derivative of $y = \text{arccot}(6x)$ at $x = 0$.
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$-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.
$-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.
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Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.
Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.
Tap to reveal answer
$\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.
$\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.
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Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.
Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.
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No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.
No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.
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Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.
Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.
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No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.
No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.
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What is the domain of $y = \text{arcsin}(x)$?
What is the domain of $y = \text{arcsin}(x)$?
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$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
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What is the domain of $y = \text{arccos}(x)$?
What is the domain of $y = \text{arccos}(x)$?
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$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
$[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.
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Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.
Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.
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$2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.
$2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.
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State the formula for the derivative of an inverse function.
State the formula for the derivative of an inverse function.
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$(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.
$(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.
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What is the derivative of $y = \text{arctan}(x)$?
What is the derivative of $y = \text{arctan}(x)$?
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$\frac{1}{1+x^2}$. Standard derivative formula for inverse tangent function.
$\frac{1}{1+x^2}$. Standard derivative formula for inverse tangent function.
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What is the derivative of $y = \text{arccot}(x)$?
What is the derivative of $y = \text{arccot}(x)$?
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$ -\frac{1}{1+x^2} $. Standard derivative formula for inverse cotangent function.
$ -\frac{1}{1+x^2} $. Standard derivative formula for inverse cotangent function.
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Which inverse function has a derivative of $-\frac{1}{1+x^2}$?
Which inverse function has a derivative of $-\frac{1}{1+x^2}$?
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$\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.
$\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.
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Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.
Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.
Tap to reveal answer
$\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.
$\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.
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Which inverse function has a derivative of $\frac{1}{1+x^2}$?
Which inverse function has a derivative of $\frac{1}{1+x^2}$?
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$\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.
$\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.
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