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(a) Let $T_{1}(x)$ be the linearization of $f(x), T_{2}(x)$ the linearization of $g(x),$ and $T(x)$ the linearization of $F(x)=f(x)+g(x),$ each near $x_{0}$. Show that $T(x)=T_{1}(x)+T_{2}(x).$(b) Let $f(x)=(1+2 x)^{1 / 3}+(1+3 x)^{1 / 4} .$ Use (a) and the previous exercise to quickly find the linearization of $f$ near $x_{0}=0.$

$$\text { (b) } T=2+\frac{17}{12} x$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

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Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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$1-4$ Find the linearizati…

For this problem, we are asked to calculate the linear ization L of X equals F of zero plus F. Prime of zero times X. Four F of X equals 1/3 plus two X at C equal zero. So to begin, we want to evaluate F of zero here, you can see that that would be equal to 1/3 plus zero. So that's just simply 1/3. Then we need to differentiate our function. In this case we'll have to apply the chain rule. We have a two that gets brought out front. Then we'd be multiplying that by negative 1/3 plus two X. All squared. Which then means that we have that F prime of zero will be equal to negative 2/9. So we have that are linear ization here L of X will be equal to 1/3 minus 2/9 X.

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