# From the local linearizations of e^x and \sin x \ \text{near} \ x = 0 , write down...

## Question:

From the local linearizations of {eq}e^x {/eq} and {eq}\sin x \ \text{near} \ x = 0 {/eq}, write down the local linearization of the function {eq}e^x \sin x {/eq}. From this result, write down the derivative of {eq}e^x \sin x \ \text{at} \ x = 0 {/eq}. Using this technique, write down the derivative of {eq}e^x \sin \left( \frac{x}{1 + x} \right) \ \text{at} \ x = 0 {/eq}.

## Linearization:

The method of linearization allows us to handle difficult expressions for functions by approximating the function value at an input. To perform the linearization, the following formula is necessary: {eq}f(x) = f(x_0) + h \cdot f'(x_{0}) {/eq}, where {eq}f(x)_0 {/eq}, {eq}h = x - x_0 {/eq}, and {eq}f'(x_{0}) {/eq} are necessary components in order to determine the unknown {eq}f(x) {/eq} value.

For {eq}f(x) = e^x {/eq}:

The approximation of {eq}f(x) = e^x {/eq} will be based off when {eq}f(0) {/eq} in each situation.

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}(e^x) \\ &= e^x \\ \end{align*} {/eq}

Now all the components to compute {eq}f(0) {/eq} are found, so simply apply the linear approximation formula to find {eq}f(0) {/eq}.

{eq}\begin{align*} L(x) = f(0)+h\cdot f'(0) &= e^0+x\cdot e^0 \\ &= 1+x\cdot 1 \\ &= x+1 \\ \end{align*} {/eq}

For {eq}f(x) = \sin(x) {/eq}:

The approximation of {eq}f(x) = \sin(x) {/eq} will be based off when {eq}f(0) {/eq} in each situation.

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}(\sin(x)) \\ &= \cos(x) \\ \end{align*} {/eq}

Now all the components to compute {eq}f(0) {/eq} are found, so simply apply the linear approximation formula to find {eq}f(0) {/eq}.

{eq}\begin{align*} L(x) = f(0)+h\cdot f'(0) &= \sin(0)+x\cdot \cos(0) \\ &= 0+x\cdot 1 \\ &= x \\ \end{align*} {/eq}

To find the local linearization of the function {eq}f(x) = e^x \sin x {/eq}, multiply the local linearizations for {eq}f(x) = e^x {/eq} and {eq}f(x) = \sin(x) {/eq}

{eq}\begin{align*} \text{ Linearization for } e^x \sin(x) &= \text{ Linearization of } e^x \cdot \text{ Linearization of } \sin(x) \\ &= (x+1) \cdot x \\ &= x^2+x \\ \end{align*} {/eq}

From the linearization given, we can now find the derivative of {eq}e^x \sin(x) {/eq}

{eq}\begin{align*} \text{ Derivative of } e^x \sin(x) \text{ Linearization derivative } &= \frac{d}{dx}(x^2+x) \\ &= 2\cdot x^{2-1}+1 \\ &= 2x+1 \\ \end{align*} {/eq}

At {eq}x = 0 {/eq}:

{eq}2(0)+1 = 0+1 = 0 {/eq}

To find the derivative of {eq}e^x \sin(\frac{x}{1+x}) {/eq} using the linearization technique, the linearization of {eq}\sin(\frac{x}{1+x}) {/eq} is needed to be calculated.

For {eq}f(x) = \sin(\frac{x}{1+x}) {/eq}:

The approximation of {eq}f(x) = \sin(\frac{x}{1+x}) {/eq} will be based off when {eq}f(0) {/eq} in each situation.

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}(\sin(\frac{x}{1+x})) \\ &= \frac{d}{dx}(\sin(\frac{x}{1+x}))\frac{d}{dx}\left(\frac{x}{1+x}\right) \\ &= \cos(\frac{x}{1+x})(\frac{\frac{d}{dx}\left(x\right)\left(1+x\right)-\frac{d}{dx}\left(1+x\right)x}{\left(1+x\right)^2}) \\ &= \cos(\frac{x}{1+x})(\frac{1\left(1+x\right)-(0+1)x}{\left(1+x\right)^2}) \\ &= \cos(\frac{x}{1+x})(\frac{1+x-x}{\left(1+x\right)^2}) \\ &= \cos(\frac{x}{1+x})(\frac{1}{\left(1+x\right)^2}) \\ \end{align*} {/eq}

Now all the components to compute {eq}f(0) {/eq} are found, so simply apply the linear approximation formula to find {eq}f(0) {/eq}.

{eq}\begin{align*} L(x) = f(0)+h\cdot f'(0) &= (\sin(\frac{0}{1+0})+x\cdot (\cos(\frac{0}{1+0})(\frac{1}{\left(1+0\right)^2})) \\ &= (\sin(\frac{0}{1})+x\cdot (\cos(\frac{0}{1})(\frac{1}{\left(1\right)^2})) \\ &= \sin(0)+x\cdot (\cos(0)(\frac{1}{1})) \\ &= 0+x\cdot (1(1)) \\ &= x \\ \end{align*} {/eq}

{eq}\begin{align*} \text{ Linearization for } e^x \sin(\frac{x}{1+x}) &= \text{ Linearization of } e^x \cdot \text{ Linearization of } \sin(\frac{x}{1+x}) \\ &= (x+1) \cdot x \\ &= x^2+x \\ \end{align*} {/eq}

From the linearization given, we can now find the derivative of {eq}e^x \sin(\frac{x}{1+x}) {/eq}

{eq}\begin{align*} \text{ Derivative of } e^x \sin(\frac{x}{1+x}) \text{ Linearization derivative } &= \frac{d}{dx}(x^2+x) \\ &= 2\cdot x^{2-1}+1 \\ &= 2x+1 \\ \end{align*} {/eq}

At {eq}x = 0 {/eq}:

{eq}2(0)+1 = 0+1 = 0 {/eq}