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A circle has equation x^2+y^2+6x+10y-7=0. Find the equation of the tangent line through the point on the circle (-8,-1).

The circle can be written (x+3) 2 +(y+5)^2 = 41 and so has center (-3,-5). The gradient through the center and the point (-8,-1) is;m = (-5+1)/(-3+8) = -4/5. The tangent line is perpendicular to this so has ...
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Answered by Rhianna L. • Maths tutor
1405 Views

Solve the equation (2x-1)/3 + (x+2)/2 +x/6 = 8

The first thing you want to do is to get rid of the fractions,Multiplying everything by 6 will help you do this You must multiply every fraction by 6 and also everything you do to the left hand side, you mus...
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Answered by Niamh M. • Maths tutor
6289 Views

given that f(x) = x^4 + 2x, find f'(x)

This is a simple differentiation question where we just differentiate by taking the power of the x and bringing that in front as a constant and then taking 1 off the power. So for the first term, x^4, we bri...
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Answered by Sheheryar F. • Maths tutor
4961 Views

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 ...
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Answered by Callum G. • Maths tutor
5754 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/...
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Answered by Henry T. • Maths tutor
4701 Views