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Differentiate [ x.ln(x)] with respect to x

The product rule is used to differentiate this since we are trying to differentiate the product of 2 parts--x and ln(x)So using the product rule which is d/dx=u.(dv/dx) +v.(du/dx)let u=x and v=ln(x)then du/d...
OL
Answered by Omolola L. • Maths tutor
4980 Views

Find the area enclosed by the curve y = 3x - x^2 and the x-axis

Start with finding limits by setting 3x - x^2 = 0, then factorise x(3 - x) = 0. Therefore x = 0 or 3. The area is the integral of 3x - x^2 between x = 0 and 3, sub in 3 and 0 into 3(x^2)/2 - (x^3)/3, which g...
SB
Answered by Sam B. • Maths tutor
18789 Views

Integrate 2x/[(x+1)(2x-4)

2x/[(x+1)(2x-4)] = [A/(x+1) + B/(2x-4)]x = A(x-2) + B(x+1) x = -1 -1 = A(-3)A = 1/3 x = 2 2 = B(3)B = 2/3 therefore.... 2x/[(x+1)(2x-4)] = [(1/3)/(x+1) + (2/3)/(x-2)]Using the identification method of integr...
LA
Answered by Lukas A. • Maths tutor
4776 Views

Prove that 1+2+...+n = n(n+1)/2 for all integers n>0. (Hint: Use induction.)

Let us procede by induction: First case: n=1. Then LHS (left hand side) = 1 and RHS (right hand side) = 1(1+1)/2 = 1. Therefore, we see that the statement is true for n=1. Now, we carry out the inductive ste...
AT
Answered by Aran T. • Maths tutor
4929 Views

Given that x = ln(sec(2y)) find dy/dx

x = ln (sec (2y)) The chain rule states that d/dy f (g (y)) = f'(g(y)). g'(y) Here g(y) = sec(2y) so g'(y) = 2.sec(2y).tan(2y) And f(y) = ln (y) so f'(y) = 1 / y Thus dx/dy = (1 / sec(2y)) . (2.sec(2y).tan(2...
DH
Answered by Dom H. • Maths tutor
13678 Views