Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let xx and yy be functions of tt with y=tanxy = \tan x. If dxdt=5\frac{dx}{dt} = 5, what is dydt\frac{dy}{dt} when x=π4x = \frac{\pi}{4}?\newlineWrite an exact, simplified answer.

Full solution

Q. Let xx and yy be functions of tt with y=tanxy = \tan x. If dxdt=5\frac{dx}{dt} = 5, what is dydt\frac{dy}{dt} when x=π4x = \frac{\pi}{4}?\newlineWrite an exact, simplified answer.
  1. Identify Relationship and Differentiate: Identify the relationship and differentiate using the chain rule.\newlineGiven y=tan(x)y = \tan(x), differentiate both sides with respect to tt.\newlinedydt=sec2(x)dxdt\frac{dy}{dt} = \sec^2(x) \cdot \frac{dx}{dt}
  2. Differentiate with Respect to t: Substitute the given values.\newlinedxdt=5\frac{dx}{dt} = 5 and x=π4x = \frac{\pi}{4}.\newlinedydt=sec2(π4)5\frac{dy}{dt} = \sec^2\left(\frac{\pi}{4}\right) \cdot 5
  3. Substitute Given Values: Calculate sec2(π4)\sec^2(\frac{\pi}{4}).
    sec(x)=1cos(x)\sec(x) = \frac{1}{\cos(x)}, so sec(π4)=1cos(π4)=1(2/2)=2\sec(\frac{\pi}{4}) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{(\sqrt{2}/2)} = \sqrt{2}.
    sec2(π4)=(2)2=2\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2.
  4. Calculate sec2(π4):\sec^2(\frac{\pi}{4}): Final calculation of dydt\frac{dy}{dt}.
    dydt=2×5=10\frac{dy}{dt} = 2 \times 5 = 10

More problems from Write and solve direct variation equations