Question Statement

Determine $f^{\prime }(x)$ for the following functions:

(a) $f(x)=x^2\left(x^3+5\right)$

(b) $f(x)=(x+9)(x-9)$

(c) $f(x)={\left(x^2+x^3\right)}^3$

(d) $f(x)=-3x^{-8}+2\sqrt{x}$

(e) $f(x)=a{x}^3+b{x}^2+cx+d$ (where $a,b,c,d$ are constants)

(f) $f(x)=x^{24}+2x^{1/2}+3x^8+9x^4$

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Background and Explanation

This question tests your ability to differentiate various algebraic functions using the power rule, chain rule, and algebraic simplification techniques. Before differentiating, it's often helpful to expand or rewrite functions into simpler forms (such as converting radicals to fractional exponents or recognizing special product patterns).

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Solution

Part (a)

First, expand the expression by distributing $x^2$:

$f(x)=x^2\left(x^3+5\right)=x^5+5x^2$

Now differentiate term by term with respect to $x$ using the power rule $\frac{d}{dx}(x^n)=n{x}^{n-1}$:

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}\left(x^5\right)+5\frac{d}{dx}\left(x^2\right)\\ & =5x^4+5(2x)\\ & =5x^4+10x\end{array}$

Part (b)

Recognize this as a difference of squares $(a+b)(a-b)=a^2-b^2$:

$f(x)=(x+9)(x-9)=x^2-81$

Differentiate with respect to $x$ (note that the derivative of a constant is zero):

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}(81)\\ & =2x-0\\ & =2x\end{array}$

Part (c)

This requires the chain rule: $\frac{d}{dx}[f(x){]}^n=n[f(x){]}^{n-1}\cdot f^{\prime }(x)$

Starting with: $f(x)={\left(x^2+x^3\right)}^3$

Apply the chain rule:

$\begin{array}{rl}{f}^{\prime }(x)& =3{\left(x^2+x^3\right)}^{3-1}\cdot \frac{d}{dx}\left(x^2+x^3\right)\\ & =3{\left(x^2+x^3\right)}^2\cdot \left(2x+3x^2\right)\\ & =3\left(2x+3x^2\right){\left(x^2+x^3\right)}^2\end{array}$

Part (d)

Rewrite $\sqrt{x}$ as $x^{1/2}$ to apply the power rule: $f(x)=-3x^{-8}+2x^{1/2}$

Differentiate term by term:

$\begin{array}{rl}{f}^{\prime }(x)& =-3\frac{d}{dx}\left(x^{-8}\right)+2\frac{d}{dx}\left(x^{1/2}\right)\\ & =-3\left(-8x^{-9}\right)+2\left(\frac{1}{2}{x}^{-1/2}\right)\\ & =24x^{-9}+x^{-1/2}\\ & =\frac{24}{x^9}+\frac{1}{\sqrt{x}}\end{array}$

Part (e)

Differentiate term by term, treating constants $a,b,c,d$ as coefficients:

$\begin{array}{rl}{f}^{\prime }(x)& =a\frac{d}{dx}\left(x^3\right)+b\frac{d}{dx}\left(x^2\right)+c\frac{d}{dx}(x)+\frac{d}{dx}(d)\\ & =a\left(3x^2\right)+b(2x)+c(1)+0\\ & =3a{x}^2+2bx+c\end{array}$

Part (f)

Differentiate each term separately using the power rule:

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}\left(x^{24}\right)+2\frac{d}{dx}\left(x^{1/2}\right)+3\frac{d}{dx}\left(x^8\right)+9\frac{d}{dx}\left(x^4\right)\\ & =24x^{23}+2\left(\frac{1}{2}{x}^{-1/2}\right)+3\left(8x^7\right)+9\left(4x^3\right)\\ & =24x^{23}+x^{-1/2}+24x^7+36x^3\\ & =24x^{23}+24x^7+36x^3+\frac{1}{\sqrt{x}}\end{array}$

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$ for any real number $n$
• Chain Rule: $\frac{d}{dx}[f(x){]}^n=n[f(x){]}^{n-1}\cdot f^{\prime }(x)$
• Difference of Squares: $(a+b)(a-b)=a^2-b^2$ (used to simplify before differentiating)
• Constant Multiple Rule: $\frac{d}{dx}[c\cdot f(x)]=c\cdot f^{\prime }(x)$
• Sum/Difference Rule: $\frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)$
• Constant Rule: $\frac{d}{dx}(c)=0$ for any constant $c$
• Radical to Exponent Conversion: $\sqrt{x}=x^{1/2}$ and $\frac{1}{x^n}=x^{-n}$

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Summary of Steps

1. Simplify first: Expand products (part a), apply difference of squares (part b), or rewrite radicals as exponents (parts d, f) to make differentiation easier
2. Apply the power rule: Multiply each term by its exponent, then reduce the exponent by 1
3. Use chain rule for composite functions (part c): Differentiate the outer function, keep the inner function, then multiply by the derivative of the inner function
4. Handle constants: Keep constant coefficients (including parameters $a,b,c,d$) during differentiation; constants alone become zero
5. Simplify final answer: Convert negative exponents back to fractions if needed for clarity