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Chapter 8: Problem 4

Differentiate the function \(f\) where \(f(x)\) is (a) \((a x+b)(c x+d)\) (b) \((a x+b) /(c x+d)\) (c) \(3 a x^{2}+5 b x+c\) (d) \(a x^{2} f(b x+c)\)

Short Answer

Expert verified
(a) \( 2acx + (ad + bc) \); (b) \( \frac{ad - bc}{(cx+d)^2} \); (c) \( 6ax + 5b \); (d) use the product rule.

Step by step solution

01

Expand the Expression for Part (a)

First, expand \( (a x + b)(c x + d) \). Using the distributive property, we have \( f(x) = a c x^2 + a d x + b c x + b d \).
02

Differentiate Part (a)

Differentiate \( f(x) = a c x^2 + (a d + b c)x + b d \) with respect to \( x \). The derivative is \( f'(x) = 2 a c x + (a d + b c) \).
03

Apply the Quotient Rule for Part (b)

For \( f(x) = \frac{a x + b}{c x + d} \), use the quotient rule: \( f'(x) = \frac{(c x + d) \cdot a - (a x + b) \cdot c}{(c x + d)^2} \). Simplifying, we get \( f'(x) = \frac{ad - bc}{(c x + d)^2} \).
04

Differentiate the Polynomial for Part (c)

Differentiate \( f(x) = 3 a x^2 + 5 b x + c \). The derivative is \( f'(x) = 6 a x + 5 b \), as the derivative of a constant \( c \) is 0.
05

Use the Product Rule for Part (d)

For \( f(x) = a x^2 f(b x + c) \), apply the product rule: \( f'(x) = a \cdot 2x \cdot f(b x + c) + a x^2 \cdot f'(b x + c) \cdot b \).
06

Conclusion

Summarizing, the derivatives are:(a) \( f'(x) = 2 a c x + (a d + b c) \). (b) \( f'(x) = \frac{ad - bc}{(c x + d)^2} \). (c) \( f'(x) = 6 a x + 5 b \). (d) \( f'(x) = a \cdot 2x \cdot f(b x + c) + a x^2 \cdot f'(b x + c) \cdot b \).

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Key Concepts

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

Distributive Property
The distributive property is a fundamental mathematical principle used to simplify expressions. It involves multiplying a single term by two or more terms inside a set of parentheses. In simple terms, it allows you to "distribute" a term across others.
  • In expression form: For any real numbers \(a, b, c\), the distributive property states that \(a(b + c) = ab + ac\).

  • This property is particularly useful when expanding polynomial expressions.

  • In the example \((a x + b)(c x + d)\), we apply distribution as follows: \((a x + b)(c x + d)\) becomes \(a c x^2 + a d x + b c x + b d\).
Whenever using distribution, make sure to multiply each term in the first parenthesis with each term in the second parenthesis separately.
Product Rule
The product rule is a differentiation technique used when dealing with the derivatives of products of two functions. If there are two functions, \(u(x)\) and \(v(x)\), their product can be differentiated using:
  • The formula: \((u \, v)' = u' \, v + u \, v'\).

  • This means you multiply the derivative of the first function \(u'\) by the second function \(v\), then add the product of \(u\) and the derivative of \(v'\).

  • In practice, consider \(f(x) = a x^2 f(b x + c)\). Here, \(u(x) = a x^2\) and \(v(x) = f(b x + c)\). By applying the product rule: \(f'(x) = a \, 2x \, f(b x + c) + a x^2 \, f'(b x + c) \, b\).
This allows us to find the derivative of a product of functions efficiently.
Quotient Rule
The quotient rule is applied to differentiate functions that are divided by each other. It comes in handy when you have a ratio of two differentiable functions:
  • If \(u(x)\) and \(v(x)\) are two functions, then the derivative of their quotient \(\left(\frac{u}{v}\right)'\) is given by \(\frac{u'v - uv'}{v^2}\).

  • It translates to "bottom times derivative of the top minus top times derivative of the bottom, all over the bottom squared."
For example, when differentiating \(f(x) = \frac{a x + b}{c x + d}\), use:
  • Top \(u(x) = a x + b\) and bottom \(v(x) = c x + d\).

  • Applying the rule, we get the derivative as \(f'(x) = \frac{(c x + d) \cdot a - (a x + b) \cdot c}{(c x + d)^2}\), which simplifies to \(f'(x) = \frac{ad - bc}{(c x + d)^2}\).
Polynomial Differentiation
Differentiating polynomials is one of the most straightforward differentiation processes. A polynomial is an expression involving sums of powers of one or more variables multiplied by coefficients.
  • The power rule is frequently used for polynomial differentiation. It states that for \(x^n\), the derivative is \(nx^{n-1}\).

  • Applying this rule makes it easier to find the derivative of each term in a polynomial.
For example, consider the polynomial \(f(x) = 3 a x^2 + 5 b x + c\):
  • The derivative of each term is computed: \(3 a x^2\) becomes \(6 a x\), \(5 b x\) becomes \(5 b\), and the constant \(c\) becomes 0.

  • Thus, the derivative \(f'(x)\) is \(6 a x + 5 b\).
This approach ensures a quick and systematic way to differentiate polynomial expressions.

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Most popular questions from this chapter

Evaluate the definite integrals (a) \(\int_{2}^{3} \frac{x \mathrm{~d} x}{\sqrt{(x+1)}}\) (b) \(\int_{0}^{1} x(x-1)^{11} d x\) (c) \(\int_{1}^{2}\left(x^{3 / 2}-\frac{1}{x^{2}}\right) \mathrm{d} x\) (d) \(\int_{0}^{\pi / 2} \sin x \mathrm{~d} x\) (e) \(\int_{0}^{2} \frac{d x}{\sqrt{\left(3+2 x-x^{2}\right)}}\) (Hint: Replace the \(x\) in (a) by \((x+1)-1\) and in (b) by \((x-1)+1 .)\)

Differentiate the function \(f\) where \(f(x)\) is (a) \((2 x+4)^{7}(3 x-2)^{5}\) (b) \((5 x+1)^{3}(3-2 x)^{4}\) (c) \(\left(\frac{1}{2} x+2\right)^{2}(x+3)^{4}\) (d) \(\left(x^{2}+x+1\right)^{2}\left(x^{3}+2 x^{2}+1\right)^{4}\) (e) \(\left(x^{5}+2 x+1\right)^{3}\left(2 x^{2}+3 x-1\right)^{4}\) (f) \((2 x+1)^{3}(7-x)^{5}\) (g) \(\left(x^{2}+4 x+1\right)(3 x+1)^{5}\)

Use Simpson's rule with \(h=0.1\) to estimate $$ \int_{0}^{1} \sqrt{\left(1+x^{3}\right)} d x $$ (Notice that by this method you have no way of knowing how accurate your estimate is.)

Differentiate the function \(f\) where \(f(x)\) is (a) \((5 x+3)^{9}\) (b) \((4 x-2)^{7}\) (c) \((1-3 x)^{6}\) (d) \(\left(3 x^{2}-x+1\right)^{3}\) (e) \(\left(4 x^{3}-2 x+1\right)^{6}\) (f) \(\left(1+x-x^{4}\right)^{5}\)

Show that \(t=\tan \frac{1}{2} x\) implies $$ \begin{aligned} &\sin x=\frac{2 t}{1+t^{2}} \\ &\cos x=\frac{1-t^{2}}{1+t^{2}} \end{aligned} $$ and $$ \mathrm{d} x=\frac{2}{1+t^{2}} \mathrm{~d} t $$ Hence integrate (a) \(\operatorname{cosec} x\) (b) \(\sec x\) (c) \(\frac{1}{3+4 \sin x}\) (d) \(\frac{1}{5 \sin x+12 \cos x}\)
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