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Whats the inverse of y = 2x+1/x-1 ?

To find the inverse we're wanting to make x the subject. First start by multiplying both sides by (x-1), this gives us y(x-1) = 2x+1. Expand the brackets giving yx - y = 2x+1. Gather all the x components ont...
AS
Answered by Amelia S. • Maths tutor
20005 Views

Differentiate with respect to x and write in its simpliest form, Y=(2x-3)/x^2?

Y=(2x-3)/x 2 Quotient Rule: dY/dx(u/v)=(u'v-v'u)/v 2 dY/dx=(2x 2 -2x(2x-3))/x 4 =(6x-2x 2 )/x 4 (Collect numerator x terms.) =(6-2x)/x 3 (Cancel the fraction by a factor of x.) dY/dx=(6-2x)/x 3
JD
Answered by Joe D. • Maths tutor
7123 Views

[FP2] Solve: 3 cosh x - 4 sinh x = 7

First write cosh and sinh as exponentials, from their respective definitions:sinh x = 1/2 (e x - e -x )cosh x = 1/2 (e x + e -x ​)So: 3 * 1/2 (e x + e -x ​) - 4 * 1/2 (e x - e -x ) = 7Multiply by 2: 3 * (e x...
WU
Answered by William U. • Maths tutor
22008 Views

How can we calculate the derivative of function f(x)= (x+2)/(x-1)?

In order to calculate the derivative of the given function, we need to apply the laws of derivation. Therefore: f '(x) = ((x+2)/(x-1))' f '(x) = [(x+2)' * (x-1) - (x+2) * (x-1)' ]/ (x-1) 2 f '(x) = [ 1 * (x-...
CB
Answered by Cosmin B. • Maths tutor
5705 Views

How can we solve a two-equation, two-unknown values?

If we have two equations that look like this: a1 * x + b1 * y = c1 and a2 * x + b2 * y = c2 where x,y are variables and a1,a2,b1,b2 are coeficients. then we solve it using the following method: We choose eit...
CB
Answered by Cosmin B. • Maths tutor
6014 Views