Math / Calculus

QuestionName Anahi
8. Find $\frac{d y}{d x}$ for $y=\arccos \left(\frac{x}{3}\right)$ A. $\frac{-3}{\sqrt{x^{2}-9}}$ B. $\frac{-1}{\sqrt{9-x^{2}}}$ C. $\frac{-x}{\sqrt{9-x^{2}}}$ D. $\frac{3}{\sqrt{9-x^{2}}}$
9. If $f(x)=e^{5 x}$ , then $\left(f^{-1}\right)^{\prime}(x)=$ A. $\frac{1}{5 x}$ B. $\frac{1}{e^{5 x}}$ C. $\frac{1}{5 e^{5 x}}$ D. $-\frac{2}{e^{5 x}}$
Use the table below to answer question 10. \begin{tabular}{|c|c|c|c|c|c|c|} \hline $x$ & $F(x)$ & $F^{\prime}(x)$ & $F^{\prime \prime}(x)$ & $G(x)$ & $G^{\prime}(x)$ & $G^{\prime \prime}(x)$ \\ \hline 2 & 6 & 4 & 7 & -5 & 3 & 10 \\ \hline 5 & 2 & -4 & -2 & -3 & 2 & -4 \\ \hline \end{tabular}
10. If $H(x)=G(F(x))$ , then $H^{\prime}(5)=$ A. 40 B. -6 C. -14 D. -12

Studdy Solution

STEP 1

1. We are given three separate problems to solve.
2. For problem 8, we need to find the derivative of an inverse trigonometric function.
3. For problem 9, we need to find the derivative of the inverse function of an exponential function.
4. For problem 10, we need to use the chain rule to find the derivative of a composite function.

STEP 2

1. Solve problem 8: Differentiate the inverse trigonometric function.
2. Solve problem 9: Differentiate the inverse of the exponential function.
3. Solve problem 10: Use the chain rule to find the derivative of the composite function.

STEP 3

To find $\frac{d y}{d x}$ for $y = \arccos\left(\frac{x}{3}\right)$ , use the chain rule. The derivative of $\arccos(u)$ is $-\frac{1}{\sqrt{1-u^2}}$ , where $u = \frac{x}{3}$ .

STEP 4

Differentiate $u = \frac{x}{3}$ with respect to $x$ :
$\frac{d u}{d x} = \frac{1}{3}$

STEP 5

Apply the chain rule:
$\frac{d y}{d x} = -\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}} \cdot \frac{1}{3} = -\frac{1}{3\sqrt{1-\frac{x^2}{9}}}$
Simplify the expression:
$\frac{d y}{d x} = -\frac{1}{\sqrt{9-x^2}}$
The correct answer is B.

STEP 6

To find $(f^{-1})'(x)$ for $f(x) = e^{5x}$ , use the formula for the derivative of the inverse function: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

STEP 7

Differentiate $f(x) = e^{5x}$ :
$f'(x) = 5e^{5x}$

STEP 8

Apply the inverse function derivative formula:
$(f^{-1})'(x) = \frac{1}{5e^{5(f^{-1}(x))}}$
Since $f(f^{-1}(x)) = x$ , we have:
$(f^{-1})'(x) = \frac{1}{5e^{5x}}$
The correct answer is C.

STEP 9

To find $H'(5)$ for $H(x) = G(F(x))$ , use the chain rule: $H'(x) = G'(F(x)) \cdot F'(x)$ .

STEP 10

Evaluate $F(5)$ using the table:
$F(5) = 2$

STEP 11

Evaluate $G'(F(5))$ and $F'(5)$ using the table:
$G'(2) = 3, \quad F'(5) = -4$

STEP 12

Apply the chain rule:
$H'(5) = G'(F(5)) \cdot F'(5) = 3 \cdot (-4) = -12$
The correct answer is D.
The solutions are:
8. B: $\frac{-1}{\sqrt{9-x^{2}}}$
9. C: $\frac{1}{5 e^{5 x}}$
10. D: $-12$

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Studdy Solution

STEP 1

STEP 2

1. Solve problem 8: Differentiate the inverse trigonometric function.
2. Solve problem 9: Differentiate the inverse of the exponential function.
3. Solve problem 10: Use the chain rule to find the derivative of the composite function.

STEP 3

To find $\frac{d y}{d x}$ for $y = \arccos\left(\frac{x}{3}\right)$ , use the chain rule. The derivative of $\arccos(u)$ is $-\frac{1}{\sqrt{1-u^2}}$ , where $u = \frac{x}{3}$ .

STEP 4

Differentiate $u = \frac{x}{3}$ with respect to $x$ :
$\frac{d u}{d x} = \frac{1}{3}$

STEP 5

Apply the chain rule:
$\frac{d y}{d x} = -\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}} \cdot \frac{1}{3} = -\frac{1}{3\sqrt{1-\frac{x^2}{9}}}$
Simplify the expression:
$\frac{d y}{d x} = -\frac{1}{\sqrt{9-x^2}}$
The correct answer is B.

STEP 6

To find $(f^{-1})'(x)$ for $f(x) = e^{5x}$ , use the formula for the derivative of the inverse function: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

STEP 7

Differentiate $f(x) = e^{5x}$ :
$f'(x) = 5e^{5x}$

STEP 8

Apply the inverse function derivative formula:
$(f^{-1})'(x) = \frac{1}{5e^{5(f^{-1}(x))}}$
Since $f(f^{-1}(x)) = x$ , we have:
$(f^{-1})'(x) = \frac{1}{5e^{5x}}$
The correct answer is C.

STEP 9

To find $H'(5)$ for $H(x) = G(F(x))$ , use the chain rule: $H'(x) = G'(F(x)) \cdot F'(x)$ .

STEP 10

Evaluate $F(5)$ using the table:
$F(5) = 2$

STEP 11

Evaluate $G'(F(5))$ and $F'(5)$ using the table:
$G'(2) = 3, \quad F'(5) = -4$

STEP 12

Apply the chain rule:
$H'(5) = G'(F(5)) \cdot F'(5) = 3 \cdot (-4) = -12$
The correct answer is D.
The solutions are:
8. B: $\frac{-1}{\sqrt{9-x^{2}}}$
9. C: $\frac{1}{5 e^{5 x}}$
10. D: $-12$

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Studdy Solution

STEP 1

STEP 2

1. Solve problem 8: Differentiate the inverse trigonometric function.2. Solve problem 9: Differentiate the inverse of the exponential function.3. Solve problem 10: Use the chain rule to find the derivative of the composite function.

STEP 3

To find dydx\frac{d y}{d x}dxdy​ for y=arccos⁡(x3)y = \arccos\left(\frac{x}{3}\right)y=arccos(3x​), use the chain rule. The derivative of arccos⁡(u)\arccos(u)arccos(u) is −11−u2-\frac{1}{\sqrt{1-u^2}}−1−u2​1​, where u=x3u = \frac{x}{3}u=3x​.

STEP 4

Differentiate u=x3u = \frac{x}{3}u=3x​ with respect to xxx:dudx=13\frac{d u}{d x} = \frac{1}{3}dxdu​=31​

STEP 5

STEP 6

To find (f−1)′(x)(f^{-1})'(x)(f−1)′(x) for f(x)=e5xf(x) = e^{5x}f(x)=e5x, use the formula for the derivative of the inverse function: (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}(f−1)′(x)=f′(f−1(x))1​.

STEP 7

Differentiate f(x)=e5xf(x) = e^{5x}f(x)=e5x:f′(x)=5e5xf'(x) = 5e^{5x}f′(x)=5e5x

STEP 8

STEP 9

To find H′(5)H'(5)H′(5) for H(x)=G(F(x))H(x) = G(F(x))H(x)=G(F(x)), use the chain rule: H′(x)=G′(F(x))⋅F′(x)H'(x) = G'(F(x)) \cdot F'(x)H′(x)=G′(F(x))⋅F′(x).

STEP 10

Evaluate F(5)F(5)F(5) using the table:F(5)=2F(5) = 2F(5)=2

STEP 11

Evaluate G′(F(5))G'(F(5))G′(F(5)) and F′(5)F'(5)F′(5) using the table:G′(2)=3,F′(5)=−4G'(2) = 3, \quad F'(5) = -4G′(2)=3,F′(5)=−4

STEP 12

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1. Solve problem 8: Differentiate the inverse trigonometric function.
2. Solve problem 9: Differentiate the inverse of the exponential function.
3. Solve problem 10: Use the chain rule to find the derivative of the composite function.

To find $\frac{d y}{d x}$ for $y = \arccos\left(\frac{x}{3}\right)$ , use the chain rule. The derivative of $\arccos(u)$ is $-\frac{1}{\sqrt{1-u^2}}$ , where $u = \frac{x}{3}$ .

Differentiate $u = \frac{x}{3}$ with respect to $x$ :
$\frac{d u}{d x} = \frac{1}{3}$

To find $(f^{-1})'(x)$ for $f(x) = e^{5x}$ , use the formula for the derivative of the inverse function: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

Differentiate $f(x) = e^{5x}$ :
$f'(x) = 5e^{5x}$

To find $H'(5)$ for $H(x) = G(F(x))$ , use the chain rule: $H'(x) = G'(F(x)) \cdot F'(x)$ .

Evaluate $F(5)$ using the table:
$F(5) = 2$

Evaluate $G'(F(5))$ and $F'(5)$ using the table:
$G'(2) = 3, \quad F'(5) = -4$