/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 \(y=\log _{e^{t}}(x-2)^{2}\) \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 15

\(y=\log _{e^{t}}(x-2)^{2}\) \(y=\frac{2 \log (x-2)}{\log x}\) $y^{\prime}=\frac{2(\log x) \times \frac{1}{x-2}-2 \log (x-2) \times \frac{1}{x}}{(\log x)^{2}}$ \(y^{\prime}=\frac{2}{\log 3}\) at \(x=3\)

Short Answer

Expert verified
**Question:** Find the first derivative of the function \(y = \log _{e^{t}}(x-2)^{2}\) and simplify it. Then, find the value of this derivative at \(x = 3\). **Answer:** The first derivative of the function is \(y' = \frac{2}{t} \times \frac{1}{(x-2)}\). The value of the first derivative at \(x = 3\) is \(y'(3) = \frac{2}{t}\).

Step by step solution

01

Simplify the function

First, we have to simplify the given function using logarithm properties. To do that, we need to use the change of base formula for logarithms: \(\log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}}\), where \(c\) can be any base. In our case, we want to use the natural logarithm (base \(e\)), so our \(c=e\). The given function is: $$ y = \log _{e^{t}}(x-2)^{2} $$ Applying the change of base formula, we have: $$ y = \frac{\log_{e}{(x-2)^{2}}}{\log_{e}{e^{t}}} $$ Now since the base of logarithm and module are the same we can rewrite it as: $$ y = \frac{2 \log_{e}{(x-2)}}{t} $$
02

Differentiate the function with respect to x

Now, we need to differentiate y with respect to x. To do that, we will need to apply the chain rule and the rules for the derivative of logarithms: $$ \frac{d}{dx}(\log_{e}(u)) = \frac{u'}{u} $$ Differentiating y with respect to x: $$ \frac{d}{dx}y = \frac{2}{t} \times \frac{d}{dx}(\log_{e}(x-2)) - 0 $$ Applying the chain rule: $$ y' = \frac{2}{t} \times \frac{(x-2)'}{(x-2)} = \frac{2}{t} \times \frac{1}{(x-2)} $$
03

Evaluate the first derivative at the given point \(x = 3\)

Now, we need to find the value of the first derivative of y at the point \(x = 3\). To do that, simply substitute this value into the expression for \(y'\): $$ y'(3) = \frac{2}{t} \times \frac{1}{(3-2)} $$ So, the value of the first derivative of \(y\) at \(x = 3\) is: $$ y'(3) = \frac{2}{t} $$

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Most popular questions from this chapter

$\begin{aligned} &f(x)=\left|\begin{array}{lll} \cos x & \sin x & \cos x \\ \cos 2 x & \sin 2 x & 2 \cos 2 x \\ \cos 3 x & \sin 3 x & 2 \cos 3 x \end{array}\right| \\ &f^{\prime}\left(\frac{\pi}{2}\right)=\left|\begin{array}{ccc} -1 & 1 & 0 \\ 0 & 0 & -2 \\ +3 & -1 & 0 \end{array}\right| \\ &+\left|\begin{array}{ccc} 0 & 0 & 0 \\ -1 & -2 & -2 \\ 0 & 0 & 0 \end{array}\right|+\left|\begin{array}{rrr} 0 & 1 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 6 \end{array}\right| \\ &=1 \end{aligned}$

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

Column-I (A) If \(y=3 e^{2 x}+2 e^{3 x}\) and \(\frac{d^{2} y}{d x^{2}}+\) a. $\frac{d y}{d x}+b y=0\(. where a and \)\mathrm{b}$ are real numbers, then \(\mathrm{a}+\mathrm{b}=\) (B) $\lim _{x \rightarrow 0^{+}}\left((x \cos x)^{x}+(x \sin x)^{1 / x}\right)=$ (C) If $\mathrm{f}(\mathrm{x})=\mathrm{x}^{\sin x}+(\sin \mathrm{x})^{\cos \mathrm{x}}\(, then \)\mathrm{f}^{\prime}\left(\frac{\pi}{2}\right)$ (D) Number of positive integer values of \(\mathrm{x}>4\) and satisfying the inequality \(\sin ^{-1}(\sin 5)<4 x-x^{2}+2\) is Column-II (P) \(\frac{\pi}{2}\) (Q) \(-1\) (R) 0 (S) 1

\(f(x)=\ln \sin x\) \(f^{\prime}(x)=\frac{1}{\ln \sin x} \times \frac{1}{\sin x} \times \cos x\) \(f^{\prime}\left(\frac{\pi}{6}\right)=\frac{-1}{\ln 2} \times \sqrt{3}\)

Assertion (A) : If a differentiable function \(f(x)\) satisfies the relation $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}-2)=0 \forall \mathrm{x} \in \mathrm{R}$, and if $\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x})\right)_{\mathrm{x}=\mathrm{a}}=\mathrm{b}$, then $\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x})\right)_{\mathrm{a}+4000}=\mathrm{b}$. Reason \((\mathbf{R}): \mathrm{f}(\mathrm{x})\) is a periodic function with period 4 .
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