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Chapter 3: Problem 25

Find the derivative of the following functions. $$y=(3 t-1)(2 t-2)^{-1}$$

Short Answer

Expert verified
Question: Determine the derivative of the function $$y=(3t-1)(2t-2)^{-1}$$ with respect to t. Answer: The derivative of the given function y with respect to t is: $$\frac{dy}{dt} = -6(2t-2)^{-3} + 3(2t-2)^{-1}$$.

Step by step solution

01

Identify u(t) and v(t)

In the given function, we can set $$u(t) = 3t-1$$ and $$v(t) = (2t-2)^{-1}$$ Now that we have identified u(t) and v(t), we will find their derivatives. Step 2: Find the derivatives of u(t) and v(t)
02

Find the derivatives of u(t) and v(t)

To find the derivatives, we have $$\frac{du}{dt} = 3$$ for u(t) since it is a linear function. For v(t), we need to apply the chain rule, which states that if we have a composition of functions, say w(t) = g(h(t)), then the derivative dw/dt is given by $$\frac{dw}{dt} = \frac{dg}{dh} \frac{dh}{dt}$$ Applying the chain rule with g(t) = t^{-1} and h(t) = 2t-2, we have $$\frac{dv}{dt} = -1(2t-2)^{-2}\cdot 2 = -2(2t-2)^{-2}$$ Step 3: Apply the product rule
03

Apply the product rule

Now, using the derivatives found in step 2, we can apply the product rule to find dy/dt: $$\frac{dy}{dt} = u\frac{dv}{dt} + v\frac{du}{dt} = (3t-1)\Big(-2(2t-2)^{-2}\Big) + (2t-2)^{-1}(3)$$ Step 4: Simplify the result
04

Simplify the result

Now, let's simplify the expression obtained in step 3. $$\frac{dy}{dt} = -6(2t-2)^{-1}(2t-2)^{-2} + 3(2t-2)^{-1}$$ Multiplying the first term by (2t-2)/(2t-2), we get $$\frac{dy}{dt} = -6(2t-2)^{-3} + 3(2t-2)^{-1}$$ Now, the derivative of the given function y is: $$\frac{dy}{dt} = \boxed{-6(2t-2)^{-3} + 3(2t-2)^{-1}}$$

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Most popular questions from this chapter

Suppose \(y=L(x)=a x+b\) (with \(a \neq 0\) ) is the equation of the line tangent to the graph of a one-to-one function \(f\) at \(\left(x_{0}, y_{0}\right) .\) Also, suppose that \(y=M(x)=c x+d\) is the equation of the line tangent to the graph of \(f^{-1}\) at \(\left(y_{0}, x_{0}\right)\) a. Express \(a\) and \(b\) in terms of \(x_{0}\) and \(y_{0}\) b. Express \(c\) in terms of \(a,\) and \(d\) in terms of \(a, x_{0},\) and \(y_{0}\) c. Prove that \(L^{-1}(x)=M(x)\)

The Witch of Agnesi The graph of \(y=\frac{a^{3}}{x^{2}+a^{2}},\) where \(a\) is a constant, is called the witch of Agnesi (named after the 18th-century Italian mathematician Maria Agnesi). a. Let \(a=3\) and find an equation of the line tangent to \(y=\frac{27}{x^{2}+9}\) at \(x=2\) b. Plot the function and the tangent line found in part (a).

Calculate the following derivatives using the Product Rule. $$\begin{array}{lll} \text { a. } \frac{d}{d x}\left(\sin ^{2} x\right) & \text { b. } \frac{d}{d x}\left(\sin ^{3} x\right) & \text { c. } \frac{d}{d x}\left(\sin ^{4} x\right) \end{array}$$ d. Based upon your answers to parts (a)-(c), make a conjecture about \(\frac{d}{d x}\left(\sin ^{n} x\right),\) where \(n\) is a positive integer. Then prove the result by induction.

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli)

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=\frac{x}{x+5}$$
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