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Differentiate with respect to x: x*cos(x)

Firstly, x cos(x) is a product of two functions of x. Therefore we can use the product rule to work out the derivative of the whole function. Differentiating each part makes it easier to visualize the formul...
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Answered by Stefan S. • Maths tutor
3631 Views

Prove cosec2A-cot2A=tanA

Cosec2A - cot2A= tanA Left hand side=1/sin2A - cos2A/sin2A =(1- cos2A)/sin2A =(1-(1- 2 sin^2⁡ A)/ 2sinAcosA =(1-1 + ( 2 sin^2⁡ A))/ 2sinAcosA =sinA/cosA =tanA Therefore left hand side of the equation is equa...
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Answered by Sherin T. • Maths tutor
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A curve is defined by parametric equations: x = t^(2) + 2, and y = t(4-t^(2)). Find dy/dx in terms of t, hence, define the gradient of the curve at the point where t = 2.

dy/dx = (dy/dt)/(dx/dt) y = t(4-t 2 ), then using differentiation of y with respect to t, dy/dt = 4 - 3t 2 x = t 2 + 2, then using differentiation of x with respect to t, dx/dt = 2t Find dy/dx by dividing dy...
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Answered by Micah W. • Maths tutor
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Express 3sin(2x) + 5cos(2x) in the form Rsin(2x+a), R>0 0<a<pi/2

Start by expanding out Rsin(2x+a) using the addition formula for sin, sin(A+B) = sin(A)cos(B)+cos(A)sin(B). Substituting 2x = A and a= b, we get that Rsin(2x+a) = R(sin(2x)cos(a) + cos(2x)sin(a)) = Rcos(a)si...
MF
Answered by Michael F. • Maths tutor
10329 Views

Solve the simultaneous equations: y=x+1, x^2+y^2=13

We already have an expression for y, so we can substitute this in:x 2 +(x+1) 2 =x 2 +(x+1)(x+1) = x 2 +x 2 +2x+1=2x 2 +2x+1 and hence 2x 2 +2x+1=13 and so 2x 2 +2x-12=0Now look for common factors. Here we ca...
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Answered by Lauren C. • Maths tutor
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