Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx}(9x + 2)^5 = 5(9x + 2)^4 \quad ?$

The missing expression is ☐.

Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx} e^{x^5 + 8} = e^{x^5 + 8} \quad ?$

The missing expression is ☐.

Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{1}{x^5 + 8} \quad ?$

The missing expression is ☐.

Question

Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx}(9x + 2)^5 = 5(9x + 2)^4 \quad ?$

The missing expression is ☐.

Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx} e^{x^5 + 8} = e^{x^5 + 8} \quad ?$

The missing expression is ☐.

Replace **?** with an expression that will make the equation valid.

$\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{1}{x^5 + 8} \quad ?$

The missing expression is ☐.

StudyXY · Accepted Answer

: To make the first equation valid, we need to find the derivative of $(9x + 2)^5$ using the chain rule. : Let $f(u) = u^5$ and $g(x) = 9x + 2$. : Applying the chain rule, we get $\frac{d}{dx}(9x + 2)^5 = \frac{d}{du} u^5 \cdot \frac{d}{dx} (9x + 2)$. : Substitute $u = 9x + 2$ back into the equation: $\frac{d}{dx}(9x + 2)^5 = 5(9x + 2)^4 \cdot 9$. : Simplify the equation: $\frac{d}{dx}(9x + 2)^5 = 45(9x + 2)^4$. : Now, compare this result with the original equation $\frac{d}{dx}(9x + 2)^5 = 5(9x + 2)^4 \quad ?$. --- Step 1: To make the second equation valid, we need to find the derivative of $e^{x^5 + 8}$ using the chain rule. Step 2: Let $f(u) = e^u$ and $g(x) = x^5 + 8$. Then, $f'(u) = e^u$ and $g'(x) = 5x^4$. Step 3: Applying the chain rule, we get $\frac{d}{dx} e^{x^5 + 8} = \frac{d}{du} e^u \cdot \frac{d}{dx} (x^5 + 8)$. Step 4: Substitute $u = x^5 + 8$ back into the equation: $\frac{d}{dx} e^{x^5 + 8} = e^{x^5 + 8} \cdot 5x^4$. Step 5: Comparing this result with the original equation $\frac{d}{dx} e^{x^5 + 8} = e^{x^5 + 8} \quad ?$, we can see that the original equation is missing a factor of $5x^4$. --- Step 1: To make the third equation valid, we need to find the derivative of $\ln \left(x^5 + 8\right)$ using the chain rule. Step 2: Let $f(u) = \ln(u)$ and $g(x) = x^5 + 8$. Then, $f'(u) = \frac{1}{u}$ and $g'(x) = 5x^4$. Step 3: Applying the chain rule, we get $\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{d}{du} \ln(u) \cdot \frac{d}{dx} (x^5 + 8)$. Step 4: Substitute $u = x^5 + 8$ back into the equation: $\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{1}{x^5 + 8} \cdot 5x^4$. Step 5: Simplify the equation: $\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{5x^4}{x^5 + 8}$. Step 6: Comparing this result with the original equation $\frac{d}{dx} \ln \left(x^5 + 8\right) = \frac{1}{x^5 + 8} \quad ?$, we can see that the original equation is missing the factor $\frac{5x^4}{x^5 + 8}$.

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QuestionAnatomy and Physiology

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Step 1
: To make the first equation valid, we need to find the derivative of 1$ using the chain rule.

Step 2
: Let 1$.

Final Answer

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QQuestionAnatomy and Physiology

AnswerHelpful

Full Solution Locked

Step 1: To make the first equation valid, we need to find the derivative of 1$ using the chain rule.

Step 2: Let 1$.

Final Answer

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QuestionAnatomy and Physiology

Answer

Step 1
: To make the first equation valid, we need to find the derivative of 1$ using the chain rule.

Step 2
: Let 1$.