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Chapter 3: Problem 43

Evaluate the derivative of the following functions at the given point. $$y=1 /(t+1) ; t=1$$

Short Answer

Expert verified
Answer: The derivative of the function at the point $$t=1$$ is $$\frac{-1}{4}$$.

Step by step solution

01

Identify the function and the point

We are given the function: $$y = \frac{1}{t+1}$$ and the point at which we need to find the derivative: $$t = 1$$
02

Find the derivative of the function dy/dt

We can rewrite the given function as: $$y = (t+1)^{-1}$$ Using the power rule, where if $$y = t^n$$ then $$\frac{dy}{dt} = nt^{n-1}$$, we can find the derivative: $$\frac{dy}{dt} = -1(t+1)^{-2} * \frac{d(t+1)}{dt}$$ Simplifying, we have: $$\frac{dy}{dt} = -1(t+1)^{-2} * 1$$ $$\frac{dy}{dt} = \frac{-1}{(t+1)^2}$$
03

Plug in the given point t=1 into the derivative

Now that we have the derivative, we can find its value at $$t=1$$: $$\frac{dy}{dt} = \frac{-1}{(1+1)^2}$$ $$\frac{dy}{dt} = \frac{-1}{2^2}$$
04

Simplify the expression to find the value of the derivative

We can simplify the expression to get the value of the derivative at the given point (t=1): $$\frac{dy}{dt} = \frac{-1}{4}$$ Therefore, the derivative of the function $$y=\frac{1}{t+1}$$ at the point $$t=1$$ is $$\frac{-1}{4}$$.

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