Question

find an equation of the tangent line to the graph of the given function at the specified point. f(x)=3e^x cos(x), (0, 3) y = resources read it submit answer 6. -/2 points a graphing calculator is recommended. let y = 8x sin(x). (a) find an equation of the tangent line to the curve y at the point (π/2, 4π). y =

Explanation:

Step1: Find the derivative of $f(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 3e^{x}$ and $v=\cos(x)$. The derivative of $u = 3e^{x}$ is $u^\prime=3e^{x}$, and the derivative of $v=\cos(x)$ is $v^\prime=-\sin(x)$. Then $f^\prime(x)=3e^{x}\cos(x)-3e^{x}\sin(x)$.

Step2: Evaluate the derivative at $x = 0$

Substitute $x = 0$ into $f^\prime(x)$. $f^\prime(0)=3e^{0}\cos(0)-3e^{0}\sin(0)$. Since $e^{0}=1$, $\cos(0)=1$ and $\sin(0)=0$, we have $f^\prime(0)=3\times1\times1 - 3\times1\times0=3$.

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 $(x_{1},y_{1})=(0,3)$ and $m = f^\prime(0)=3$. Substituting these values, we get $y - 3=3(x - 0)$.

Step4: Simplify the equation

$y-3 = 3x$, so $y=3x + 3$.

Answer:

$y = 3x+3$