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Find the derivatives of the functions in Exercises \(19-38\) $$ p=\sqrt{3-t} $$

Short Answer

Expert verified
The derivative is \( \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \).

Step by step solution

01

Identify the function

The function given is \( p = \sqrt{3-t} \). This can be rewritten as \( p = (3-t)^{1/2} \) to make it easier to apply differentiation rules.
02

Apply the Chain Rule

To differentiate \( p = (3-t)^{1/2} \), we need to apply the chain rule. The chain rule states that \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). Here, \( f(x) = x^{1/2} \) and \( g(t) = 3-t \).
03

Differentiate the outer function

The outer function is \( f(x) = x^{1/2} \). The derivative \( f'(x) = \frac{1}{2}x^{-1/2} \). So, when applying \( g(t) \), \( f'(g(t)) = \frac{1}{2}(3-t)^{-1/2} \).
04

Differentiate the inner function

The inner function \( g(t) = 3-t \) has a derivative of \( g'(t) = -1 \).
05

Apply the Chain Rule Derivative

Now apply the chain rule: \[ \frac{dp}{dt} = \frac{1}{2}(3-t)^{-1/2} \cdot (-1) \]. This simplifies to \( \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \).
06

Write the final derivative

The derivative of \( p = \sqrt{3-t} \) is \( \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule in Differentiation
The chain rule is a powerful differentiation tool. It is used when you have to differentiate a composite function—essentially a function inside another function. This is common in calculus, especially with functions that are not directly differentiable using simpler rules.
For instance, consider the function given in the exercise: \( p = \sqrt{3-t} \). When rewritten, it becomes \( p = (3-t)^{1/2} \). To find its derivative with respect to \( t \), we apply the chain rule. The chain rule helps by breaking down the differentiation process into manageable steps.
To apply the chain rule, identify your outer function \( f(x) \) and your inner function \( g(t) \). For our example, \( f(x) = x^{1/2} \) and \( g(t) = 3-t \). The chain rule states that the derivative of a composite function \( f(g(t)) \) is \( f'(g(t)) \cdot g'(t) \). This approach makes differentiation systematic and your task easier.
  • Outer Function: Differentiate the outer function.
  • Inner Function: Differentiate the inner function.
  • Combine: Multiply these derivatives together to get the final derivative of the composite function.
Understanding Derivatives
Derivatives are a fundamental instrument in calculus. They measure how a function changes as its input changes. Simply put, a derivative tells you the slope of a function at any point. Slopes inform you whether a function is increasing, decreasing, or constant at certain points.
When taking the derivative of a function like \( p = \sqrt{3-t} \), we effectively find how \( p \) changes with respect to \( t \). Differentiation rules such as the power rule, product rule, and specifically the chain rule, guide this process.
In mathematical terms, for our function \( p = \sqrt{3-t} \), the derivative is \( \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \). This represents the rate of change of \( p \) as \( t \) changes. The negative sign indicates that for increasing values of \( t \), \( p \) is decreasing.
Performing Function Analysis
Function analysis involves examining a function to understand its behavior. This involves derivatives, which provide insight into slope and rate of change, but also entails more than just differentiation.
For \( p = \sqrt{3-t} \), analyzing this function starts with understanding its domain and range, which tells where the function is defined and what values it can take.
Using the derivative, \( \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \), we can further analyze its behavior:
  • Sign of the Derivative: Indicates where the function increases or decreases.
  • Critical Points: Points where the derivative equals zero or is undefined, which can signal maxima, minima, or points of inflection.
  • Concavity: Derivatives also help in determining the concavity of a function, which tells you how the function bends and if it contains any turning points.
Function analysis using derivatives is crucial for creating graphs, solving real-world problems, and deeper comprehension of function traits.

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Most popular questions from this chapter

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\cos t, \quad y=\sqrt{3} \cos t, \quad t=2 \pi / 3 $$

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the volume \(V=\pi r^{2} h\) of a right circular cylinder when the radius changes from \(r_{0}\) to \(r_{0}+d r\) and the height does not change

Estimate the allowable percentage error in measuring the diameter \(D\) of a sphere if the volume is to be calculated correctly to within 3\(\%\) .

a. Given that \((x-2)^{2}+y^{2}=4\) find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions with respect to \(x\) and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \((x-2)^{2}+y^{2}=4\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \((x-2)^{2}+y^{2}=4 .\) Then add the graphs of the functions' first derivatives to your picture. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \((x-2)^{2}+y^{2}=4 ?\) Could you have predicted the general behavior of the graph of \((x-2)^{2}+y^{2}=4\) by looking at the derivative graphs? Give reasons for your answers.

The folium of Descartes (See Figure 3.38\()\) a. Find the slope of the folium of Descartes, \(x^{3}+y^{3}-9 x y=0\) at the points \((4,2)\) and \((2,4) .\) b. At what point other than the origin does the folium have a horizontal tangent? c. Find the coordinates of the point \(A\) in Figure \(3.38,\) where the folium has a vertical tangent.

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