Question

Find the first and second derivatives of f(x) 1-2x O f'(x) 2x-3 ex 5 - 2x f"(x) = O 3-2 f'(x) = 2x-5 F"(x) ex O f'(x) = 4xe* - 2e f"(x) = 4xe* + 2e O f'(x) = 2xe* - 3e* f'(x) = 2xe - ex

Answer 1

Given the function

$f\mleft(x\mright)=\frac{1-2x}{e^x}^{}$

To compute the first derivative, we must recall the derivative of a quotient rule:

$(\frac{h}{g})^{\prime}=\frac{h^{\prime}g-g^{\prime}h}{g^2}$

Here: h(x)=1-2x, g(x)=e^x

h'(x)=-2

g'(x)=e^x

substituting into the formula:

$f^{\prime}(x)=\frac{(-2)e^x-e^x(1-2x)^{}}{e^{2x}}^{}=\frac{-2e^x-e^x+2xe^x}{e^{2x}}$

Operating:

$f^{\prime}(x)=\frac{-3e^x+2xe^x}{e^{2x}}=e^x(\frac{-3+2x}{e^{2x}})$

Simplifying numerator and denominator:

$f^{\prime}(x)=\frac{-3+2x}{e^x}$

Looks like this matches the first choice

Now find the second derivative

This time: h(x)=-3+2x, h'(x)=2

g(x)=e^x, g'(x)=e^x

Applying the quotient rule again:

$f^{\prime}^{\prime}(x)=\frac{2e^x-e^x(-3+2x)}{e^{2x}}=\frac{2e^x+3e^x-2xe^x}{e^{2x}}=\frac{5e^x-2xe^x}{e^{2x}}$

Factoring and simplifying:

$f^{\doubleprime}(x)=e^x\cdot\frac{5-2x}{e^{2x}}=\frac{5-2x}{e^x}$

Both answers confirm the first choice is correct