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Find the derivative of the following functions. $$f(x)=\frac{2 e^{x}-1}{2 e^{x}+1}$$

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{2e^x-1}{2e^x+1}\) is \(f'(x) = \frac{4e^x}{(2e^x+1)^2}\).

Step by step solution

01

Identify the functions g(x) and h(x)

In our expression, we have g(x) in the numerator and h(x) in the denominator: $$g(x)=2e^{x}-1$$ $$h(x)=2e^{x}+1$$ Now that we have g(x) and h(x), we need to find their derivatives.
02

Find the derivatives g'(x) and h'(x)

To find the derivatives, we'll apply the chain rule. $$g'(x) = \frac{d}{dx} (2e^{x}-1) = 2\frac{d}{dx}e^{x} = 2e^{x}$$ $$h'(x) = \frac{d}{dx} (2e^{x}+1) = 2\frac{d}{dx}e^{x} = 2e^{x}$$
03

Apply the quotient rule

Now we can apply the quotient rule to find the derivative of f(x): $$f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{\left[h(x)\right]^2}=\frac{(2e^{x})(2e^{x}+1)-(2e^{x}-1)(2e^{x})}{\left[2e^{x}+1\right]^2}$$
04

Simplify

Now we have to simplify the expression for the derivative: $$f'(x) =\frac{4e^{2x}+2e^{x}-4e^{2x} + 2e^{x}}{\left[2e^{x}+1\right]^2}=\frac{4e^{x}}{\left[2e^{x}+1\right]^2}$$ So, the function's derivative is: $$f'(x)=\frac{4e^{x}}{(2e^{x}+1)^2}$$

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Most popular questions from this chapter

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

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Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

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