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When you differentiate an equation, you're finding the gradient of its graph.<o:p></o:p> For example, if you differentiate the equation y = x 2 you get a solution dy/dx = 2x<o:p></o:p&gt...

If y is a function of u, which itself is a function of x, then dy/dx=(dy/du) x (du/dx) Differentiate the outer function and multiply by the derivative of the inner function. To illustrate this rule, look at ...

let y=2 x {take natural logs of both sides} ln y = ln(2 x ) {use rules of logs to change right hand side} lny = xln2 {differentiate implicitly} 1/y . dy/dx = ln2 {make dy/dx the subject} dy/ dx = y ln2 {writ...

cos2x= 2cos^2x - 1 2cos^2x - 1 - 5cosx = 2 Let cosx = y 2y^2 - 5y - 3 = 0 (2y + 1)(y + 3) = 0 y = -1/2 or -3 cosx = -1/2 or -3 x = 120 degrees

= 1 + 3x 1/3 + (1/3 -2/3)(3x)^2 + (1/3*-2/3*-5/3)(3x)^3 = 1 + x - 2x^2 + 10x^3