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Find the differential and evaluate for the given xx and dxd x. Answer exactly. y=tan(x),x=7π4,dx=π11dy=2π11\begin{array}{l} y=\tan (x), x=\frac{7 \pi}{4}, d x=\frac{\pi}{11} \\ d y=\frac{2 \pi}{11} \end{array} 0 Submit Question

Studdy Solution

STEP 1

What is this asking? We need to find the *differential* of the function y=tan(x)y = \tan(x) and then calculate its value at x=7π4x = \frac{7\pi}{4} with a tiny change in xx of dx=π11dx = \frac{\pi}{11}. Watch out! Remember that the differential is *not* the same as the derivative!
Also, be careful with trig functions and their values at different angles.

STEP 2

1. Find the Derivative
2. Calculate the Differential
3. Evaluate the Differential

STEP 3

Let's **find the derivative** of our function y=tan(x)y = \tan(x).
The derivative of tan(x)\tan(x) with respect to xx is sec2(x)\sec^2(x).
So, we have: dydx=sec2(x) \frac{dy}{dx} = \sec^2(x)

STEP 4

Now, let's **calculate the differential** dydy.
The differential dydy is defined as: dy=dydxdx dy = \frac{dy}{dx} \cdot dx We already found dydx\frac{dy}{dx} in the previous step.
Substituting that in, we get: dy=sec2(x)dx dy = \sec^2(x) \cdot dx

STEP 5

Time to **plug in the values**!
We're given x=7π4x = \frac{7\pi}{4} and dx=π11dx = \frac{\pi}{11}.
Let's substitute these values into our differential: dy=sec2(7π4)π11 dy = \sec^2\left(\frac{7\pi}{4}\right) \cdot \frac{\pi}{11}

STEP 6

Now, let's **evaluate** sec(7π4)\sec\left(\frac{7\pi}{4}\right).
Remember that sec(x)=1cos(x)\sec(x) = \frac{1}{\cos(x)}.
The cosine of 7π4\frac{7\pi}{4} is 22\frac{\sqrt{2}}{2}, so: sec(7π4)=1cos(7π4)=122=22=2 \sec\left(\frac{7\pi}{4}\right) = \frac{1}{\cos\left(\frac{7\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

STEP 7

So, sec2(7π4)=(2)2=2\sec^2\left(\frac{7\pi}{4}\right) = (\sqrt{2})^2 = 2.
Substituting this back into our equation for dydy, we get: dy=2π11=2π11 dy = 2 \cdot \frac{\pi}{11} = \frac{2\pi}{11}

STEP 8

The differential dydy is 2π11\frac{2\pi}{11}.

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