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Find the equation of the straight line tangent to the curve y=2x^3+3x^2-4x+7, at the point x=-2.

We are looking for a straight line, so it needs the form y=mx+c. To find our gradient, m, we need the gradient of the curve at the point x=-2, so differentiate the equation: dy/dx=6x 2 +6x-4, and solve at x=...
JB
Answered by James B. • Maths tutor
6533 Views

When do you use integration by parts?

The formula for integration by parts is Integral(u dv/dx)dx = uv - Integral(v du/dx)dx You use integration by parts when you have an integral where you have to terms multiplied together ie Integral(u*dv/dx)d...
AB
Answered by Alex B. • Maths tutor
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The curve C has equation 16*y^3 + 9*x^2*y - 54*x = 0 a)Find dy/dx in terms of x and y

16y 3 + 9x 2 y - 54x= 0 a) Differentiate the terms separately dy/dx(16y 3 + 9x 2 y - 54x) = dy/dx(16y 3 ) + dy/dx(9x 2 y) - dy/dx(54x) = 48y 2 (dy/dx) + 18xy + 9x 2 (dy/dx) - 54 Implicit differentiation, tre...
JG
Answered by Joseph G. • Maths tutor
4850 Views

Integrate x/(x^2+2)

0.5Ln(x^2+2)+ C
RS
Answered by Rajan S. • Maths tutor
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f (x) = (x^2 + 4)(x^2 + 8x + 25). Find the roots of f (x) = 0

firstly, x 2 + 4 = 0 x 2 = -4 x = 2i x = -2iSecondly, x 2 + 8x + 25 = 0 using the quadratic formulae: x = (-b +- sqrt(b 2 - 4ac))/2a x = (-8+-sqrt(64-100))/2 x = -8/2 +- sqrt(-36)/2 x = -4 + 3i x = -4 - 3i
LS
Answered by Laura S. • Maths tutor
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