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Maths
A Level

Find the stationary pointsof the following: (y = x^3 - x^2 -16 x -17) and determine if each point is a maximum or minimum.

Notes; *Stationary (Turning) points are the points on the graph which are lowest or highest. (maximum or minima). *The gradient at a stationary point is zero. Steps:  1. Differentiate the function once to...

Answered by Charlie M. Maths tutor
3192 Views

find the value of dy/dx at the point (1,1) of the equation e^(2x)ln(y)=x+y-2

find dy/dx, algebraically manipulate the expression to get dy/dx in terms of x and y and then subsitute in the given point. 

Answered by James G. Maths tutor
7174 Views

Given that x=ln(t) and y=4t^3,a) find an expression for dy/dx, b)and the value of t when d2y/dx2 =0.48. Give your answer to 2 decimal place.

a) Firstly, differentiate x and y with respect to t. 

Giving you dx/dt = 1/t       and dy/dt = 12t2

dy/dx is found using the chain rule:

dy/dx = dy/dt x dt/dx = 12t3

Answered by Sara W. Maths tutor
2916 Views

Calculate the gradient of the function y=x^2+6x when y=-9

-9=x^2+6x

0=x^2+6x+9

0=(x+3)(x+3)

when y=-9 x=3

dy/dx=2x+6

dy/dx=2(3)+6=12

Answered by Hassan M. Maths tutor
4643 Views

For a curve of gradient dy/dx = (2/(x^2))-x/4, determine a) d^2y/dx^2 b) the stationary point where y=5/2 c) whether this is a maximum or minmum point and d) the equation of the curve

a) Differentiating gives d2y/dx2=-4x-3-1/4

b) Let dy/dx=0 and rearrange to find x=2

c) Inserting x=2 into d2y/dx2=-4x-3-1/4 ...

Answered by Katie M. Maths tutor
5228 Views

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