Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Sep 19, 2023

Differentiate the function $y = \frac{1}{p+k e^{p}}$ and find $y'(p)$ .

STEP 1

Question: What does the derivative of a function at a point measure?
A) The rate of change of the function at that point. B) The slope of the tangent line at that point. C) Both A and B. D) None of the above.
Answer: C) Both A and B.

STEP 2

Question: How can we rewrite the function $y=\frac{1}{p+ke^{p}}$ to simplify the differentiation process?
A) $y=(p+ke^{p})^{-1}$ B) $y=p+ke^{-p}$ C) $y=p+ke^{p}$ D) $y=(p+ke^{p})^2$
Answer: A) $y=(p+ke^{p})^{-1}$

STEP 3

Question: Which rule can we apply to differentiate the function $y=(p+ke^{p})^{-1}$ ?
A) Product rule B) Quotient rule C) Chain rule D) Power rule
Answer: C) Chain rule

STEP 4

Question: If we apply the chain rule to differentiate $y=(p+ke^{p})^{-1}$ , what is the derivative?
A) $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+ke^{p})$ B) $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+ke^{-p})$ C) $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+ke^{2p})$ D) $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+k)$
Answer: A) $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+ke^{p})$

STEP 5

Question: What is the simplified form of the derivative $y^{\prime}(p) = -1 \cdot (p+ke^{p})^{-2} \cdot (1+ke^{p})$ ?
A) $y^{\prime}(p) = -\frac{1+ke^{p}}{(p+ke^{p})^{2}}$ B) $y^{\prime}(p) = \frac{1+ke^{p}}{(p+ke^{p})^{2}}$ C) $y^{\prime}(p) = -\frac{1+ke^{2p}}{(p+ke^{p})^{2}}$ D) $y^{\prime}(p) = -\frac{1+ke^{-p}}{(p+ke^{p})^{2}}$
Answer: A) $y^{\prime}(p) = -\frac{1+ke^{p}}{(p+ke^{p})^{2}}$

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