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Compute \(y^{\prime}(x)\) for a. \(\quad y=2 x^{3}-5\) b. \(y=\frac{2}{x^{2}}\) c. \(y=\frac{5}{(x+1)^{2}}\) d. \(y=\left(1+x^{2}\right)^{0.5}\) e. \(\quad y=\sqrt{1-x^{2}}\) f. \(y=\left(1-x^{2}\right)^{-0.5}\) g. \(\quad y=(2-x)^{4}\) h. \(y=\left(3-x^{2}\right)^{4}\) i. \(y=\frac{1}{\sqrt{7-x^{2}}}\) j. \(y=\left(1+(x-2)^{2}\right)^{2}\) k. \(y=(1+3 x)^{1.5} \quad\) l. \(\quad y=\frac{1}{\sqrt{16-x^{2}}}\) \(\mathrm{m} . \quad y=\sqrt{9-(x-4)^{2}}\) n. \(y=\left(9-x^{2}\right)^{1.5}\) o. \(y=\sqrt[3]{1-3 x}\)

Short Answer

Expert verified
The derivatives are: a) 6x², b) -4/x³, c) -10/(x+1)³, d) x/√(1+x²), e) -x/√(1-x²), f) x/(1-x²)^{1.5}, g) -4(2-x)³, h) -8x(3-x²)³, i) x/(7-x²)^{1.5}, j) 4(x-2)(1+(x-2)²), k) 4.5(1+3x)^{0.5}, l) x/(16-x²)^{1.5}, m) (x-4)/√(9-(x-4)²), n) -3x(9-x²)^{0.5}, o) -1/(1-3x)^{2/3}.

Step by step solution

01

Differentiate part a

The function given is \( y = 2x^{3} - 5 \). To find the derivative \( y'(x) \), apply the power rule. The derivative of \( 2x^3 \) is \( 6x^2 \) and the derivative of constant \( -5 \) is \( 0 \), so \( y'(x) = 6x^2 \).
02

Differentiate part b

For \( y = \frac{2}{x^2} = 2x^{-2} \), use the power rule. The derivative is \( y'(x) = -4x^{-3} = -\frac{4}{x^3} \).
03

Differentiate part c

For \( y = \frac{5}{(x+1)^2} = 5(x+1)^{-2} \), use the power rule. The derivative is \( y'(x) = -10(x+1)^{-3} = -\frac{10}{(x+1)^3} \).
04

Differentiate part d

Given \( y = (1+x^2)^{0.5} \), use the chain rule: \( u = 1+x^2 \), \( u' = 2x \), so \( y'(x) = 0.5(1+x^2)^{-0.5} \, * \, 2x = \frac{x}{\sqrt{1+x^2}} \).
05

Differentiate part e

For \( y = \sqrt{1-x^2} = (1-x^2)^{0.5} \), use the chain rule: \( u = 1-x^2 \), \( u' = -2x \), hence \( y'(x) = 0.5(1-x^2)^{-0.5} \, * \, (-2x) = \frac{-x}{\sqrt{1-x^2}} \).
06

Differentiate part f

Given \( y = (1-x^2)^{-0.5} \), apply the chain rule: \( u = 1-x^2 \), \( u' = -2x \), thus \( y'(x) = -0.5(1-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(1-x^2)^{1.5}} \).
07

Differentiate part g

The function is \( y = (2-x)^4 \). Using the chain rule where \( u = 2-x \) and \( u' = -1 \), we have \( y'(x) = 4(2-x)^3 \, * \, (-1) = -4(2-x)^3 \).
08

Differentiate part h

For \( y = (3-x^2)^4 \), use the chain rule: \( u = 3-x^2 \), \( u' = -2x \), so \( y'(x) = 4(3-x^2)^3 \, * \, (-2x) = -8x(3-x^2)^3 \).
09

Differentiate part i

The function is \( y = \frac{1}{\sqrt{7-x^2}} = (7-x^2)^{-0.5} \). Apply chain rule: \( u' = -2x \), so \( y'(x) = -0.5(7-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(7-x^2)^{1.5}} \).
10

Differentiate part j

Given \( y = (1+(x-2)^2)^2 \), employ the chain rule: \( u = 1+(x-2)^2 \), \( u' = 2(x-2) \), hence \( y'(x) = 2(1+(x-2)^2) \, * \, 2(x-2) = 4(x-2)(1+(x-2)^2) \).
11

Differentiate part k

For \( y = (1+3x)^{1.5} \), apply the chain rule: \( u = 1+3x \), \( u' = 3 \), thus \( y'(x) = 1.5(1+3x)^{0.5} \, * \, 3 = 4.5(1+3x)^{0.5} \).
12

Differentiate part l

The function is \( y = \frac{1}{\sqrt{16-x^2}} = (16-x^2)^{-0.5} \). Using the chain rule: \( u' = -2x \), so \( y'(x) = -0.5(16-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(16-x^2)^{1.5}} \).
13

Differentiate part m

For \( y = \sqrt{9-(x-4)^2} = (9-(x-4)^2)^{0.5} \), use chain rule: \( u = 9-(x-4)^2 \), \( u' = -2(x-4) \), so \( y'(x) = 0.5(9-(x-4)^2)^{-0.5} \, * \, (-2(x-4)) = \frac{(x-4)}{\sqrt{9-(x-4)^2}} \).
14

Differentiate part n

Given \( y = (9-x^2)^{1.5} \), apply chain rule: \( u = 9-x^2 \), \( u' = -2x \), so \( y'(x) = 1.5(9-x^2)^{0.5} \, * \, (-2x) = -3x(9-x^2)^{0.5} \).
15

Differentiate part o

The function is \( y = \sqrt[3]{1-3x} = (1-3x)^{1/3} \). Using the chain rule: \( u = 1-3x \), \( u' = -3 \), so \( y'(x) = \frac{1}{3}(1-3x)^{-2/3} \, * \, (-3) = -\frac{1}{(1-3x)^{2/3}} \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus, used when you need to differentiate composite functions. A composite function is a combination of two or more functions, where one function is nested inside another. To apply the chain rule, you differentiate the outer function, keeping the inner function intact, and then multiply it by the derivative of the inner function.

The formula for the chain rule is:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]Here, \(y\) is a function of \(u\), and \(u\) is a function of \(x\). This approach lets you tackle complex differentiation problems step-by-step, making it easier to manage.

For example, consider the function \(y = (1 + x^2)^{0.5}\). To differentiate using the chain rule:
  • Let \(u = 1 + x^2\), then \(y = u^{0.5}\).
  • The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du} = 0.5u^{-0.5}\).
  • The derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx} = 2x\).
  • Apply the chain rule: \(\frac{dy}{dx} = 0.5(1+x^2)^{-0.5} \cdot 2x = \frac{x}{\sqrt{1+x^2}}\).
This methodology is crucial when simplifying and differentiating functions, allowing us to dive deeper into the behavior of functions and their graphs.
Power Rule
The power rule is a practical tool in differentiation, making it simple to find the derivatives of functions where the variable is raised to a power. The rule states that if you have a function in the form \(y = x^n\), where \(n\) is a constant, then the derivative \(y'\) is given by \(nx^{n-1}\).

For a better understanding, consider this example: if \(y = 2x^3\), applying the power rule gives \(y' = 3 \cdot 2x^{3-1} = 6x^2\). It's straightforward and efficient, making it suitable for any polynomial expression.

  • The power rule also extends to negative and fractional exponents.
  • For instance, if \(y = x^{-2}\), then \(y' = -2x^{-3}\).
  • In fractional form, such as \(y = x^{1/2}\), the derivative becomes \(y' = \frac{1}{2}x^{-1/2}\).
This rule helps streamline the process of differentiation, allowing for quick analysis and comprehension of polynomial trends and behaviors in function graphs.
Functions and Graphs
Understanding functions and their graphs is crucial in calculus, as it helps visualize the impact of differentiation. Functions can be simple linear equations, polynomial expressions, or more complex shapes like trigonometric or exponential forms. Graphically, differentiation provides insight into the slope of the tangent lines at various points on a function's curve.

  • The derivative tells us the rate of change of the function at any given point.
  • A positive derivative indicates that the function is increasing, while a negative derivative shows it is decreasing.
  • A zero derivative points to a potential maximum or minimum value, also known as critical points.
By differentiating simple functions like polynomials or fractions, we gain valuable information about the graph's behavior such as concavity and points of inflection.

For example, in the function \(y = (2-x)^4\), differentiation helps us understand the rate and direction of change as \(x\) varies. Through these insights, one can predict how the graph will look and identify any significant characteristics like symmetry or intercepts.

Analyzing functions through their derivatives is a powerful method to explore their graphical representation and underlying mechanics.

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Most popular questions from this chapter

For the function, \(f(x)=\sin (x)\), find an an open interval, \((3-\delta, 3+\delta)\) so that \(f(x)>0\) for \(x\) in \((3-\delta, 3+\delta)\).

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Exercise \(4.1 .9 \quad\) a. Draw the graph of a function, \(f,\) defined on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) b. Does your graph intersect the X-axis? c. Draw a graph of of a function, \(f,\) defined on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) that does not intersect the X-axis. Be sure that its X-projection is all of [1,3] . d. Write equations to define a function, \(f,\) on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) and the graph of \(f\) does not intersect the X-axis. e. There is a theorem that asserts that the function you just defined must be discontinuous at some number in [1,3] . Identify such a number for your example. The preceding exercise illustrates a general property of continuous functions called the intermediate value property. Briefly it says that a continuous function defined on an interval that has both positive and negative values on the interval, must also be zero somewhere on the interval. In language of graphs, the graph of a continuous function defined on an interval that has a point below the X-axis and a point above the X-axis must intersect the X-axis. The proof of this property requires more than the familiar properties of addition, multiplication, and order of the real numbers. It requires the completion property of the real numbers, Axiom \(5.2 .1^{2}\).

For those points that are on the graph, find the slopes of the tangents to the graph of a. \(\quad \frac{x^{2}}{18}+\frac{y^{2}}{8}=1\) at the points (3,2) and (-3,2) b. \(\frac{2 x^{2}}{35}+\frac{3 y^{2}}{35}=1\) at the points \((4,1), \quad(-3,-2),\) and (4,-1)

Suppose \(p\) and \(q\) are integers and \(u\) is a positive function that has a derivative at all numbers \(t\). Assume that \(\left[u^{\frac{p}{q}}(t)\right]^{\prime}\) exists. Give reasons for the steps \((i)-(i v)\) below that show $$ \left[u^{\frac{p}{q}}(t)\right]^{\prime}=\frac{p}{q} u^{\frac{p}{q}-1}(t) \times u^{\prime}(t) $$ Let $$ v(t)=u^{\frac{p}{q}}(t) $$ Then $$ \begin{aligned} v^{q}(t) &=u^{p}(t) \\ \left[v^{q}(t)\right]^{\prime} &=\left[u^{p}(t)\right]^{\prime} \\ q v^{q-1}(t) \times v^{\prime}(t) &=p u^{p-1}(t) \times u^{\prime}(t) \\ v^{\prime}(t) &=\frac{p u^{p-1}(t)}{q} \frac{p}{\left(u^{\frac{p}{q}}\right)^{q-1}} \times u^{\prime}(t) \\ \left[u^{\frac{p}{q}}(t)\right]^{\prime} &=\frac{p}{q} u^{\frac{p}{q}-1}(t) \times u^{\prime}(t) \end{aligned} $$

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