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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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Answered by Leo R. • Maths tutor

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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...

Answered by Sarah P. • Maths tutor

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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...

Answered by Chantelle C. • Maths tutor

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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) = -(...

Answered by Andrew P. • Maths tutor

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Express 5/[(x-1)(3x+2)] as partial fractions.

5 = a(3x+2) + b(x-1), x = 1, 5 = a(3+2) +b (1-1), 5 = a(5) + b(0), 5 = 5a, a = 1, x = -2/3, 5 = a(-2+2) + b(-(2/3)-1), 5 = a(0) + b(-5/3), 5 = -5b/3, b = -3, Therefore: 5/[(x-1)(3x+2)] = 1/(x-1) -3/(3...

Answered by Andrew P. • Maths tutor

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