Kisha saved $12.00 when buying a coat. The coat was on sale for 30% off. What was the original price of the coat?
Differentiate the function.
a. g(x) – In(x VP – 1).
b. F(y) = y ln(1 + e’).
The Correct Answer and Explanation is:
Let’s break your request into two parts:
Part 1: Solving the Word Problem
Problem:
Kisha saved $12.00 with a 30% discount on a coat. What was the original price?
Solution:
Let the original price be x.
30% of x = $12
→ 0.30x=120.30x = 12
→ x=120.30=40x = \frac{12}{0.30} = 40
✅ Original price = $40.00
Part 2: Differentiating the Given Functions
Let’s differentiate each function step-by-step and explain.
a. g(x)=ln(x⋅P−1)g(x) = \ln(x \cdot \sqrt{P} – 1)
Note: It seems you meant a function like ln(xP−1)\ln(x\sqrt{P} – 1), assuming “VP” meant “√P”. Let’s denote it clearly: g(x)=ln(xP−1)g(x) = \ln(x\sqrt{P} – 1)
Use the chain rule:
- Let u=xP−1u = x\sqrt{P} – 1
- Then g(x)=ln(u)g(x) = \ln(u)
- So g′(x)=1u⋅dudxg'(x) = \frac{1}{u} \cdot \frac{du}{dx}
ddx[xP−1]=P\frac{d}{dx}[x\sqrt{P} – 1] = \sqrt{P} g′(x)=PxP−1g'(x) = \frac{\sqrt{P}}{x\sqrt{P} – 1}
✅ Answer: g′(x)=PxP−1g'(x) = \frac{\sqrt{P}}{x\sqrt{P} – 1}
b. F(y)=yln(1+ey)F(y) = y \ln(1 + e^y)
Use the product rule: F′(y)=ddy[y]⋅ln(1+ey)+y⋅ddy[ln(1+ey)]F'(y) = \frac{d}{dy}[y] \cdot \ln(1 + e^y) + y \cdot \frac{d}{dy}[\ln(1 + e^y)]
- ddy[y]=1\frac{d}{dy}[y] = 1
- To differentiate ln(1+ey)\ln(1 + e^y), use the chain rule: ddy[ln(1+ey)]=11+ey⋅ey=ey1+ey\frac{d}{dy}[\ln(1 + e^y)] = \frac{1}{1 + e^y} \cdot e^y = \frac{e^y}{1 + e^y}
Putting it all together: F′(y)=ln(1+ey)+y⋅ey1+eyF'(y) = \ln(1 + e^y) + y \cdot \frac{e^y}{1 + e^y}
✅ Answer: F′(y)=ln(1+ey)+yey1+eyF'(y) = \ln(1 + e^y) + \frac{y e^y}{1 + e^y}
Explanation (Approx. 300 Words)
Differentiation is a core concept in calculus, used to determine how a function changes as its input changes. In both problems above, we use foundational rules of differentiation: the chain rule and product rule.
In function a, we’re differentiating a natural logarithm function involving a linear term. The chain rule is essential when dealing with a composition of functions. For example, ln(u(x))\ln(u(x)) requires applying the derivative of ln first, 1u\frac{1}{u}, then multiplying it by the derivative of the inner function uu. This is why we took the derivative of xP−1x\sqrt{P} – 1 (which is P\sqrt{P}) and multiplied it by 1xP−1\frac{1}{x\sqrt{P} – 1}.
In function b, we’re differentiating a product of two functions: yy and ln(1+ey)\ln(1 + e^y). When differentiating products, we use the product rule, which says (uv)′=u′v+uv′(uv)’ = u’v + uv’. The second part of the product includes a logarithmic function of an exponential expression, which itself requires the chain rule. The derivative of ln(1+ey)\ln(1 + e^y) involves first differentiating the outer ln function, then multiplying by the derivative of the inner 1+ey1 + e^y, giving ey1+ey\frac{e^y}{1 + e^y}.
Altogether, these problems test our ability to apply differentiation rules in a layered way. Recognizing which rules to apply — chain rule, product rule, and properties of logarithmic and exponential functions — is key to solving such calculus problems correctly.