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Maths
A Level

Evaluate the following : ∫ln(x) dx

Integrate by parts:u = ln(x) u' = 1/x v' = 1 v = xBy parts formula: uv - ∫u'v dx Therefore we have: xln(x) - ∫x1/x dx = xln(x) - ∫1 dx = xln(x) - x (+c)

SM
Answered by Saaqib M. Maths tutor
6447 Views

What is the chain rule and how is it used?

The chain rule is used to find the derivative of an expression in the form h(f(x)) where you have a function in terms of a function of x for example:
h(f(x)) = 2(3x+1)^3 where f(x) = 3x+1
In or...

JB
Answered by Joseph B. Maths tutor
3273 Views

Use the addition formulas: sin(x+y)=sin(x)*cos(y)+sin(y)*cos(x), cos(x+y)=cos(x)*cos(y)-sin(x)*sin(y) to derive sin(2x), cos(2x), sin(x)+sin(y).

sin(2x)=sin(x+x) =sin(x)cos(x)+cos(x)sin(x) =2*sin(x)*cos(x); cos(2x)=cos(x+x) =cos(x)cos(x)-sin(x)sin(x) =[cos(x)]2-[sin(x)]2 Now let x+y=a and x-y=b, then x=(a+b)/2 and y=(a-b)/2. ...

IS
Answered by Ioana S. Maths tutor
4224 Views

A curve has the equation, 6x^2 +3xy−y^2 +6=0 and passes through the point A (-5, 10). Find the equation of the normal to the curve at A.

Use implicit differentiation on original equation-
12x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0
dy/dx= -12x -3y/(3x-2y) at A, x= -5 and y= 10 therefore, dy/dx=-6/7

To find the normal of t...

CG
Answered by Chantal G. Maths tutor
6245 Views

A curve has the equation y = (x^2 - 5)e^(x^2). Find the x-coordinates of the stationary points of the curve.

This requires the chain rule and the product rule to be used to differentiate the function. The substitution u = x2 can be used to make this easier. Using this, du/dx = 2x and y = (u-5)eu<...

OJ
Answered by Oliver J. Maths tutor
4082 Views

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