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If f(x) = sin(2x)/(x^2) find f'(x)

As f(x) is in the form of u(x)/v(x) we can apply the rule that f'(x) = (u'(x)*v(x) - v'(x)*u(x))/(v(x)2), pulled from the C3 formula booklet.
If u(x) = sin(2x) then u'(x) = 2cos(2x).
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LR
Answered by Leo R. Maths tutor
3609 Views

The function f (x) is defined by f (x) = (1-x)/(1+x), x not equal to -1. Show that f(f (x)) = x. Hence write down f ^-1 (x).

f(f (x) )= f( (1-x)/(1+x) ) = (1-(1-x)/(1+x))/(1+(1-x)/(1+x))where you replace x by (1-x)/(1+x). Multiply the top and bottom of the fraction by (1+x) to get ((1+x)-(1-x))/((1+x)+(1-x)) which simplifies t...

SP
Answered by Sarah P. Maths tutor
11521 Views

What is the normal distribution and how do I use it?

The normal distribution is a distribution we can use when we know the mean and the standard deviation of a population, to work out probabilities that a certain even will occur.
The main properties of...

CC
Answered by Chantelle C. Maths tutor
3392 Views

Ed has 4 cards. There is a number on each card. Three of the numbers are 12, 6 and 15. The mean of the numbers is 10. What is the fourth number?

To find the mean you add all the numbers together and divide by the number of numbers. Call the fourth number x. Then we know that (12+6+15+x)/4 =10. then multiply both sides by 4 to get 12+6+15+x=40 whi...

SP
Answered by Sarah P. Maths tutor
9464 Views

Integrate x/((1-x^2)^0.5) with respect to x

x = sin(u), dx/du = cos(u), dx = cos(u) * du,[x/(1-x^2)^0.5)] * dx = [sin(u)/((1-(sin(u)^2))^0.5] * cos(u) * du = [sin(u)/(cos(u)^2)^0.5] * cos(u) * du = sin(u) * duIntegral of sin(u) * du = -cos(u) = -(...

AP
Answered by Andrew P. Maths tutor
4403 Views

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