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Chapter 4: Problem 5

\(f\left(x^{2}\right)=x^{4}+x^{3}+1\) Put \(x=x^{2}\) \(f\left(x^{4}\right)=x^{8}+x^{6}+1\) \(f^{\prime}\left(x^{4}\right)=\frac{8 x^{7}+6 x^{5}}{4 x^{3}}\) \(=2 x^{4}+\frac{3}{2} x^{2}\)

Short Answer

Expert verified
Answer: The expression for \(f\left(x^4\right)\) is \(x^{16} + x^{12} + 1\), and its derivative \(f^{\prime}\left(x^4\right)\) is \(x^{11}(16x^4 + 12)\).

Step by step solution

01

Perform the substitution

Since we are given the substitution \(x = x^2\), we can rewrite the original function as \(f(x) = (x^2)^4 + (x^2)^3 + 1\). Now, we can substitute \(y = x^2\) into the expression as \(f(y) = y^4 + y^3 + 1\). Finally, we substitute \(y = x^4\) to obtain the expression for \(f\left(x^4\right)\).
02

Apply the substitution

Substitute \(y = x^4\) into the expression for \(f(y)\): $$ f\left(x^4\right) = (x^4)^4 + (x^4)^3 + 1 $$
03

Simplify the expression

Calculate the powers to obtain: $$ f\left(x^4\right) = x^{16} + x^{12} + 1 $$
04

Compute the derivative of \(f\left(x^4\right)\)

Find \(f^{\prime}\left(x^4\right)\) by differentiating the terms with respect to \(x\): $$ f^{\prime}\left(x^4\right) = \frac{d}{dx}(x^{16} + x^{12} + 1) $$
05

Apply the Chain Rule

Differentiate with respect to \(x\), applying the chain rule: $$ f^{\prime}\left(x^4\right) = 16x^{15} + 12x^{11} $$
06

Simplify the derivative

Factor out the common term \(x^{11}\) to simplify the expression: $$ f^{\prime}\left(x^4\right) = x^{11}(16x^4 + 12) $$ Now we have found the expression for \(f\left(x^4\right)\) and its derivative \(f^{\prime}\left(x^4\right)\).

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Most popular questions from this chapter

\(y=x^{2}\) \(\frac{d y}{d x}=2 x \quad \frac{d x}{d y}=\frac{1}{2 x}\) $\frac{d^{2} y}{d x^{2}}=2 \quad \frac{d^{2} x}{d y^{2}}=\frac{-1 \times 2}{(2 x)^{2}} \times \frac{d x}{d y}=\frac{-1}{2 x^{2}} \times \frac{1}{2 x}$ \(\frac{d^{2} y}{d x^{2}} \cdot \frac{d^{2} x}{d y^{2}}=\frac{-1}{2 x^{3}}\)

$\begin{aligned} &a x^{3}+b x^{2}+b x+d=0\\\ &3 a x^{2}+2 b x+b=0\\\ &\Rightarrow 3 \mathrm{a}+3 \mathrm{~b}=0 \quad \text { (as } 1 \text { is repeated root) }\\\ &\Rightarrow a+b=0\\\ &\text { Now, a }+b+b+d=0\\\ &\Rightarrow \mathrm{b}+\mathrm{d}=0 \end{aligned}$

\(f(x)=x^{2} \ln g(x)\) \(f^{\prime}(x)=2 x \ln g(x)+\frac{x^{2} g^{\prime}(x)}{g(x)}\) \(f^{\prime}(2)=4 \ln 3-\frac{16}{3}\)

\(y=f^{-1}(x)\) As $g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^{3}}=-\frac{4}{8}=-\frac{1}{2}$

$\begin{aligned} &y=e^{\sqrt{x}}+e^{-\sqrt{x}} \\ &\begin{array}{l} \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{2 \sqrt{x}}-\frac{e^{-\sqrt{x}}}{2 \sqrt{x}} \\ =\frac{\sqrt{\left(e^{\sqrt{x}}+e^{-\sqrt{x}}\right)^{2}-4}}{2 \sqrt{x}}=\frac{\sqrt{y^{2}-4}}{2 \sqrt{x}} \end{array} \end{aligned}$
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