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

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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 ☐.

CramX · Accepted Answer

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

Answer

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$.

Step 3:
: Applying the chain rule, we get 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$.

Step 3:: Applying the chain rule, we get 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$.

Step 3:
: Applying the chain rule, we get 1$.