Can someone help me figure out the answer and..possibly show the steps.. to finding the derivative of these 3 questions..

1. f(x)=2(e^x) - x^2

2. f(x)=e^x/(e^x +1)

3. y=10^(√x)

#1. (e^x)' = e^x, so

f' = 2e^x - 2x

#2.
quotient rule:
f' = (e^x(e^x+1) - e^x*e^x)/(e^x+1)^2
= (e^2x + e^x - e^2x)/(e^x+1)^2
= e^x/(e^x+1)^2

or, knowing that e^x/(e^x+1) = 1 - 1/(e^x+1)
f' = e^x/(e^x+1)^2

#3. Recall that (a^u)' = ln a a^u u'

y' = ln 10 10^√x * 1/(2√x) = ln10/(2√x) 10^(√x)

d/dx(1+8/x)*

Of course! I'd be happy to help you find the derivatives of these functions and walk you through the steps. Let's go through each question one by one.

1. To find the derivative of f(x) = 2(e^x) - x^2, we can use the basic rules of differentiation.

The derivative of e^x is e^x. Therefore, the derivative of 2(e^x) is simply 2(e^x).

The derivative of x^2 is 2x. Therefore, the derivative of -x^2 is -2x.

Now, to find the derivative of the entire function, we add the derivatives of both terms:

f'(x) = 2(e^x) - 2x

2. To find the derivative of f(x) = e^x / (e^x + 1), we can use the quotient rule.

The quotient rule states that for a function u(x) / v(x), the derivative is given by:

(u'(x)v(x) - v'(x)u(x)) / v(x)^2

Let u(x) = e^x and v(x) = (e^x + 1).

u'(x) = e^x (the derivative of e^x is e^x)
v'(x) = derivative of (e^x + 1) = e^x (the derivative of e^x is e^x)

Now, we can substitute these values into the quotient rule formula:

f'(x) = (e^x(e^x + 1) - e^x(e^x)) / (e^x + 1)^2

Simplifying further:

f'(x) = e^x / (e^x + 1)^2

3. To find the derivative of y = 10^(√x), we can use the chain rule.

Let's break down the function into two parts:
u(x) = 10^x (the base function)
v(x) = √x (the exponent function)

The derivative of u(x) with respect to x is 10^x times the derivative of v(x) with respect to x.

The derivative of √x is 1 / (2√x).

Therefore, the derivative of y = 10^(√x) is:

y' = 10^(√x) * (1 / (2√x))

These are the steps to calculate the derivatives of the given functions. If you have any further questions, please let me know!