Use the table to find the following derivatives. x 1 2 3 4 5 f^'(

Name: Problem 60, Chapter 3: Calculus Early Transcendentals
Uploaded: 2021-08-10T14:22:35-08:00
Duration: 1 min 39 s
Channel: Nick Johnson
Description: Use the table to find the following derivatives. x 1 2 3 4 5 f^'(x) 3 5 2 1 4 g^'(x) 2 4 3 1 5 .(d)/(d x)(2 x-3 g(x))|x=4

step 1 Hello. So here we have a given table of x values and then f prime of x and g prime of x. So what

Question

Use the table to find the following derivatives. $$\begin{array}{cccccc}x & 1 & 2 & 3 & 4 & 5 \\\hline f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\g^{\prime}(x) & 2 & 4 & 3 & 1 & 5\end{array}$$ $$\left.\frac{d}{d x}(2 x-3 g(x))\right|_{x=4}$$

   Use the table to find the following derivatives.
$$\begin{array}{cccccc}x & 1 & 2 & 3 & 4 & 5 \\\hline f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\g^{\prime}(x) & 2 & 4 & 3 & 1 & 5\end{array}$$
$$\left.\frac{d}{d x}(2 x-3 g(x))\right|_{x=4}$$

Calculus Early Transcendentals

William Briggs, Lyle… 2nd Edition

Chapter 3, Problem 60

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Solved by Expert Nick Johnson

08/10/2021

Step 1

The derivative of a function is given by the sum of the derivatives of its parts. So, we can write the derivative as: $$\frac{d}{dx}(2x - 3g(x)) = 2\frac{d}{dx}(x) - 3\frac{d}{dx}(g(x))$$ Show more…

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Use the table to find the following derivatives. $$\begin{array}{cccccc}x & 1 & 2 & 3 & 4 & 5 \\\hline f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\g^{\prime}(x) & 2 & 4 & 3 & 1 & 5\end{array}$$ $$\left.\frac{d}{d x}(2 x-3 g(x))\right|_{x=4}$$

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Key Concepts

Using Tabulated Derivative Values

This concept involves extracting derivative values from a provided table, allowing one to efficiently evaluate the derivative of a function at specified points without needing to compute them from the basic definition of the derivative.

Derivatives of Linear Functions

The derivative of a linear function, such as the term '2x' in an expression, is the constant coefficient of x. Recognizing this helps quickly determine parts of a derivative when combined with other function evaluations.

Linearity of Differentiation

This principle states that the derivative of a sum (or difference) of functions is the sum (or difference) of their derivatives, and that constants can be factored out. It is essential for breaking down and differentiating complex expressions composed of simpler functions.

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