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A curve has an equation y=3x-2x^2-x^3. Find the x-coordinate(s) of the stationary point(s) of the curve.

The very first step in solving this problem is understanding that a stationary point is where the derivative of the curve, dy/dx (or in Newton’s notation y’), is equal to zero. This is because at stationary ...
CG
Answered by Callum G. • Maths tutor
5759 Views

Use the quotient rule to differentiate: ln(3x)/(e^4x) with respect to x.

Quotient rule: d(u/v)/dx = [(du/dx)v-u(dv/dx)]/v^2 u = ln(3x) v = e^4x Find du/dx using chain rule: u = ln(z) ==> du/dz = 1/z z = 3x ==> dz/dx = 3 (du/dz)(dz/dx) = 3/z = 3/3x = 1/x du/dx = 1/x Find dv/...
HT
Answered by Henry T. • Maths tutor
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Solve the inequality x^2 – 5x – 14 > 0.

In order to solve this, the quadratic must be factorised. This means we're trying to get the equation into the form (x+a)(x+b)>0 where a and b are constants where a x b = -14 and a + b = -5. This turns ou...
MP
Answered by Marcus P. • Maths tutor
11455 Views

Differentiate y = x^2 - 2x-3 + e^3x + 2ln(x)

The above function has several components with different rules on how to differentiate each of them. For each:(1) y = x 2 dy/dx = 2x This follows the general rule of differentiating polynomials: y = x n , dy...
JH
Answered by Jim H. • Maths tutor
4191 Views

A curve is defined by the parametric equations x = 2t and y = 4t^2 + t. Find the gradient of the curve when t = 4

the gradient of the curve = dy/dx and dy/dx = (dy/dt)(dt/dx) dy/dt = 8t + 1 dx/dt = 2 therefore dt/dx = 1/2 dy/dx as above = (8t + 1) * 1/2 = (8t + 1)/2 where t = 4, dy/dx = (8*4 + 1)/2 = (32 + 1)/2 = 33/2
AB
Answered by Angus B. • Maths tutor
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