Question

differentiate the function.
y = e^{x + 8}+9
y=

resources
read it master it

submit answer

1. -/1 points

details my notes ask your teacher practice another

find an equation of the tangent line to the curve at the given point.
y = 5e^{x}+x, (0, 5)
y =

resources
read it

Explanation:

Step1: Recall derivative rules

The derivative of $e^u$ with respect to $x$ is $e^u\cdot u'$ by the chain - rule, and the derivative of a constant is 0. For the function $y = e^{x + 8}+9$, let $u=x + 8$, then $u'=1$.

Step2: Differentiate the function

The derivative of $e^{x + 8}$ is $e^{x + 8}\cdot1$ (by the chain - rule), and the derivative of 9 (a constant) is 0. So $y'=e^{x + 8}$.

for the second part:

Step1: Find the derivative of the function

The derivative of $y = 5e^{x}+x$ is $y'=5e^{x}+1$ (since the derivative of $e^{x}$ is $e^{x}$ and the derivative of $x$ is 1).

Step2: Evaluate the derivative at the given point

Substitute $x = 0$ into $y'$. When $x = 0$, $y'(0)=5e^{0}+1=5\times1 + 1=6$. This is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,5)$ and $m = 6$. Substituting these values, we get $y-5=6(x - 0)$.

Step4: Simplify the equation

$y-5=6x$, so $y = 6x+5$.

Answer:

$e^{x + 8}$