Question

consider the following equation.
y = x^4 + 4e^x
find y(x).
y(x) = 4x^3 + 4e^x
find equations of the tangent line and normal line to the given curve at the point (0, 4).
tangent line y = 4x + 4
normal line y = i

Explanation:

Step1: Recall derivative rules

The derivative of $x^n$ is $nx^{n - 1}$ and the derivative of $e^x$ is $e^x$. For $y=x^{4}+4e^{x}$, by the sum - rule of derivatives $(u + v)'=u'+v'$, where $u = x^{4}$ and $v = 4e^{x}$. The derivative of $x^{4}$ is $4x^{3}$ and the derivative of $4e^{x}$ is $4e^{x}$, so $y'(x)=4x^{3}+4e^{x}$.

Step2: Find the slope of the tangent line at $(0,4)$

Substitute $x = 0$ into $y'(x)$. $y'(0)=4(0)^{3}+4e^{0}=4$. The equation of a line in point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,4)$ and $m = 4$. So $y-4 = 4(x - 0)$, which simplifies to $y=4x + 4$.

Step3: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line $m_{t}=4$, the slope of the normal line $m_{n}=-\frac{1}{4}$. Using the point - slope form with $(x_{1},y_{1})=(0,4)$ and $m =-\frac{1}{4}$, we have $y - 4=-\frac{1}{4}(x - 0)$, which simplifies to $y=-\frac{1}{4}x + 4$.

Answer:

$y=-\frac{1}{4}x + 4$