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Find the derivative (dy/dx) of the curve equation x^2 -y^2 +y = 1.

Most of the differentiation problems require us to apply one of the well known rules, be it product rule, quotient rule or chain rule. But those problems have one thing in common: explicite formula for y, be...
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Answered by Adam G. • Maths tutor
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The equation of a line is y=3x – x^3 a) Find the coordinates of the stationary points in this curve, stating whether they are maximum or minimum points b) Find the gradient of a tangent to that curve at the point (2,4)

a) A stationary point is any point on the curve that is flat, still, not increasing or decreasing. Another way to think of this is the gradient at a stationary point = 0 Firstly make an equation for your gra...
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Answered by Lauren C. • Maths tutor
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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 fin...
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Answered by Charlie M. • Maths tutor
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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.
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Answered by James G. • Maths tutor
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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 = 12t 2 dy/dx is found using the chain rule: dy/dx = dy/dt x dt/dx = 12t 3 b) You will need to differentiate dy/dx again ...
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Answered by Sara W. • Maths tutor
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