question 3\nif $f(x)=e^{2x - 10}$\nfind the slope of the tangent line to f(x) at x = 5\n\nquestion 4\nif…

Question 3

Explanation:

Step1: Differentiate the function

Using the chain - rule, if $y = e^{u}$ and $u = 2x-10$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}(e^{u})=e^{u}$ and $\frac{d}{dx}(2x - 10)=2$, we have $f^\prime(x)=2e^{2x - 10}$.

Step2: Evaluate the derivative at $x = 5$

Substitute $x = 5$ into $f^\prime(x)$: $f^\prime(5)=2e^{2\times5-10}=2e^{0}=2\times1 = 2$.

Answer:

$2$

Question 4

Explanation:

Step1: Differentiate the function

Let $u=-4x^{2}+4x$, then $y = e^{u}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{d}{du}(e^{u})=e^{u}$ and $\frac{d}{dx}(-4x^{2}+4x)=-8x + 4$. So $f^\prime(x)=(-8x + 4)e^{-4x^{2}+4x}$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$: $f^\prime(1)=(-8\times1 + 4)e^{-4\times1^{2}+4\times1}=(-4)e^{0}=-4\times1=-4$.

Answer:

$-4$