Question

Given $f(x)=2\sec (x)$, write the equation of the line tangent to $y=f(x)$ when $x=\frac{5\pi }{3}$.
Answer Attempt 1 out of 2
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Step 1 ：Find the derivative of the function $f(x)=2\sec (x)$.

Step 2 ：The derivative of $\sec (x)$ is $\sec (x)\tan (x)$, so the derivative of $f(x)$ is $f^{\prime }(x)=2\sec (x)\tan (x)$.

Step 3 ：Evaluate the derivative at the point $x=\frac{5\pi }{3}$ to find the slope of the tangent line at that point.

Step 4 ：$f^{\prime }(\frac{5\pi }{3})=2\sec (\frac{5\pi }{3})\tan (\frac{5\pi }{3})$.

Step 5 ：We know that $\sec (\frac{5\pi }{3})=\frac{1}{\cos (\frac{5\pi }{3})}=-2$ and $\tan (\frac{5\pi }{3})=\frac{\sin (\frac{5\pi }{3})}{\cos (\frac{5\pi }{3})}=\sqrt{3}$.

Step 6 ：So, $f^{\prime }(\frac{5\pi }{3})=2\ast (-2)\ast \sqrt{3}=-4\sqrt{3}$.

Step 7 ：This is the slope of the tangent line at the point $x=\frac{5\pi }{3}$.

Step 8 ：Find the y-coordinate of the point on the function where $x=\frac{5\pi }{3}$.

Step 9 ：$f(\frac{5\pi }{3})=2\sec (\frac{5\pi }{3})=2\ast (-2)=-4$.

Step 10 ：So, the point on the function where the tangent line touches is $(\frac{5\pi }{3},-4)$.

Step 11 ：Use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is $y-y_1=m(x-x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope of the line.

Step 12 ：Substituting our values, we get $y-(-4)=-4\sqrt{3}(x-\frac{5\pi }{3})$.

Step 13 ：Simplifying, we get $y=-4\sqrt{3}x+4\sqrt{3}\ast \frac{5\pi }{3}+4$.

Step 14 ：So, the equation of the line tangent to $y=f(x)$ when $x=\frac{5\pi }{3}$ is $\boxed{{\displaystyle y=-4\sqrt{3}x+4\sqrt{3}\ast \frac{5\pi }{3}+4}}$.