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given $f(x)=4sin^{2}(x)$, write the equation of the line tangent to $y …

Question

given $f(x)=4sin^{2}(x)$, write the equation of the line tangent to $y = f(x)$ when $x=\frac{7pi}{4}$.

Explanation:

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

Using the chain - rule. Let $u = \sin(x)$, then $y = 4u^{2}$. $\frac{dy}{du}=8u$ and $\frac{du}{dx}=\cos(x)$. So $f^\prime(x)=8\sin(x)\cos(x)$.

Step2: Evaluate the derivative at $x = \frac{7\pi}{4}$

Substitute $x=\frac{7\pi}{4}$ into $f^\prime(x)$. $\sin(\frac{7\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\cos(\frac{7\pi}{4})=\frac{\sqrt{2}}{2}$. Then $f^\prime(\frac{7\pi}{4})=8\times(-\frac{\sqrt{2}}{2})\times\frac{\sqrt{2}}{2}=- 4$.

Step3: Evaluate the function at $x = \frac{7\pi}{4}$

Substitute $x = \frac{7\pi}{4}$ into $f(x)$. $f(\frac{7\pi}{4})=4\sin^{2}(\frac{7\pi}{4})=4\times(-\frac{\sqrt{2}}{2})^{2}=2$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(\frac{7\pi}{4},2)$ and $m=-4$. So $y - 2=-4(x-\frac{7\pi}{4})$.

Step5: Simplify the equation

$y-2=-4x + 7\pi$, then $y=-4x+7\pi + 2$.

Answer:

$y=-4x + 7\pi+2$