# Calculus problems

what is the derivative of the following functions with respect to x:
a) y = 5e^-4x ans: -20e ^-4x
b) y = 1/2e^x^2 ans: xe^x^2
c)y = x^4 e^x ans: 4x^3 e^x + x^4 e^x
d) y = e^-x (x) ans: - (x + 1)e^-x/(x^2)
e) y = (1 + e^x)^1/2 ans: e^x/[2(1 + e^x)^1/2]
f) y = x + e ^ (x)^1/2 ans: 1 + [e ^(x)^1/2 / 2(x)^1/2]

3a) what is the equation of the tangent to the curve at the specified
point?
a) y = e^2x at (1, e^2) ans: y = 2e^2x - e^2
b) y = e^x^2/(x) at (1, e) ans: y = x

4)a) what is d^2/y/dx^2

a) y = x^2 e^-x ans: (x^2 -4x +2)e^-x
b) y = 4xe^x^2 ans: (16 x^3 + 24 x)e^x^2
c) y = e^-x sin x ans: -2e^-x cos x

5) how do you show that if y = e^x cos 2x then
d^2y/dx - 2(dy/dx) + 5y = 0?

=====
How do you evaluate log (subcript 2) 1/32?
b) log (g) 32 + log g(16) ?
c) log (2) 3
how do you solve for x? are there restrictions?
a) log (x) 25 = 2/3
b) log(7) [x + 7] + log(7) [x-7] = 0

#### bogdan

Shouldn't this be in the Homework Help forum ?

#### FZ+

What are you stuck on? What can you do?

#### Sting

Just a hint:

A function such as f(x) = e^x, you simply multiply the derivative of the exponent with the actual function. For example:

f(x) = e^(4x)

exponent is 4x and the derivative is 4 so:

f'(x) = 4 * e^(4x) = 4e^(4x)

That's the basic tenet of derivatives involving e.

If you need anymore help, just tell us where you are having trouble.

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