Derivatives

• Jan 30th 2013, 01:00 PM
hcharrington
Derivatives
How would i solve these derivatives?

1. y=5sec(pi)x tan(pi)x

2. y=ln(e^(-x^2))

3. y= sin sqrt(x) + sqrt(sin2x)

4. y = arctanx + x/
• Jan 30th 2013, 01:33 PM
AnonymousUser
Re: Derivatives
You need to review the product rule and chain rule as you will be using them both. Do you know what they are?
Second, you need to review the rule for derivatives of trig functions and ln and e. For example .... if y = ln(u) then y' = 1/(u) * u'

Memorize these rules in general and you will be able to apply them now matter what.
• Jan 30th 2013, 02:28 PM
HallsofIvy
Re: Derivatives
Quote:

Originally Posted by hcharrington
How would i solve these derivatives?

1. y=5sec(pi)x tan(pi)x

This is written badly. What you have written is the constant "5 sec(pi)tan(pi) x^2" but I suspect you mean 5 sec(pi x)tan(pi x). Do you know the derivative of sec(x)? Do you know the derivative of tan(x)? Do you know the chain rule and product rule?

Quote:

2. y=ln(e^(-x^2))
You know that ln and exp are inverse functions, don't you? So that ln(e^{-x^2))= -x^2.

[quote]3. y= sin sqrt(x) + sqrt(sin2x)[quote]
Do you know the derivative of sin(x), and square root of x? Do you know the chain rule?

Quote:

4. y = arctanx + x/
Was there something after that "/"? The derivative of arctan(x) is 1/(1+ x^2). I will give you that one because it is not a common as the others. (I got it by googling on "derivative" and "arctangent" and got Derivatives of Inverse Trigonometric Functions.)
• Jan 30th 2013, 02:35 PM
AnonymousUser
Re: Derivatives
Quote:

Originally Posted by HallsofIvy

You know that ln and exp are inverse functions, don't you? So that ln(e^{-x^2))= -x^2

You will get the same result if you simply take the derivative of ln and then use the chain rule, which is probably good practice for taking derivatives.