mcv 4u - unit 4 test - derivatives of sinusoidal functions page 2 of 4\n2. find the exact equation of the…

Explanation:

Step1: Find the derivative of the function

The derivative of $y = 4x-\tan x$ is $y'=4-\sec^{2}x$ using the power - rule for $4x$ ($(ax)' = a$) and the derivative of $\tan x=\sec^{2}x$.

Step2: Evaluate the derivative at $x = \frac{\pi}{3}$

Substitute $x=\frac{\pi}{3}$ into $y'$. We know that $\sec x=\frac{1}{\cos x}$, and $\cos\frac{\pi}{3}=\frac{1}{2}$, so $\sec\frac{\pi}{3} = 2$. Then $y'\left(\frac{\pi}{3}\right)=4 - 2^{2}=4 - 4=0$.

Step3: Find the y - coordinate of the point

Substitute $x = \frac{\pi}{3}$ into the original function $y = 4x-\tan x$. So $y=4\times\frac{\pi}{3}-\tan\frac{\pi}{3}=\frac{4\pi}{3}-\sqrt{3}$.

Step4: Write the equation of the tangent line

The equation of a tangent line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})$ is the point of tangency and $m$ is the slope. Here $x_{1}=\frac{\pi}{3}$, $y_{1}=\frac{4\pi}{3}-\sqrt{3}$ and $m = 0$. So the equation is $y-(\frac{4\pi}{3}-\sqrt{3})=0\times(x - \frac{\pi}{3})$, which simplifies to $y=\frac{4\pi}{3}-\sqrt{3}$.

Answer:

$y=\frac{4\pi}{3}-\sqrt{3}$