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Chapter 3: Problem 55

Calculate the derivative of the following functions. $$y=4 \log _{3}\left(x^{2}-1\right)$$

Short Answer

Expert verified
Answer: The derivative of the function \(y = 4 \log_{3}(x^{2} - 1)\) with respect to \(x\) is \(\frac{dy}{dx} = \frac{8x}{(x^{2} - 1)\ln(3)}\).

Step by step solution

01

Recognize the Outer and Inner Function

We have two functions in this case: the outer function is the logarithm function (\(\log_{3}\)), and the inner function is the quadratic function (\(x^{2} - 1\)). We will use the chain rule to differentiate this composite function.
02

Differentiate the Outer Function

The derivative of the logarithm function with respect to its input is the reciprocal of the input, so the derivative of \(\log_{3}(u)\) with respect to \(u\) is \(\frac{1}{u\ln(3)}\).
03

Differentiate the Inner Function

The derivative of the quadratic function \((x^{2} - 1)\) with respect to \(x\) is \(2x\).
04

Apply the Chain Rule

To differentiate the composite function, we multiply the derivative of the outer function by the derivative of the inner function. So, we get: $$\frac{dy}{dx} = 4\cdot \frac{1}{(x^{2} - 1)\ln(3)}\cdot 2x$$
05

Simplify the Expression

We can simplify the expression by multiplying the constants and combining terms: $$ \frac{dy}{dx} = \frac{8x}{(x^{2} - 1)\ln(3)}$$ So the derivative of the function \(y = 4 \log_{3}(x^{2} - 1)\) with respect to \(x\) is: $$\frac{dy}{dx} = \frac{8x}{(x^{2} - 1)\ln(3)}$$

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Most popular questions from this chapter

An observer is \(20 \mathrm{m}\) above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is \(20 \mathrm{m}\) horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of \(5 \mathrm{m} / \mathrm{s}\), what is the rate of change of the angle of elevation when the elevator is \(10 \mathrm{m}\) above the ground? When the elevator is \(40 \mathrm{m}\) above the ground?

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=x^{2}-2 x-3, \text { for } x \leq 1 ;(12,-3)$$

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of $$f(x)=x^{3} ;(8,2)$$

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