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Unlike a straight line, the gradient of a curve is not a constant i.e. not one single number. To find the gradient of a curve, you different the equation of the curve. To find the gradient at a specific p...

First, we choose u=sin(x),v'=exp(x). Using differentiation and integration of standard exponential and trigonometric functions => u'=cos(x),v=exp(x). From this we use the formula for integration by par...

Differentiate by multiply by the power and then minus 1 from the power. Use the rule: y= ax^b -> dy/dx = abx^(b-1)

dy/dx = 3 x 2x^(3-1) + -1 x 2x^(-1-1) + 0 dy/dx = 6x^2 - 2x^-2

Use the method of integration by parts. uv-integral(v.du/dx). Make u equal to ln(x) and dv/dx equal to 1. Therefore v=x and du/dx=1/x. Hence uv=xln(x). And v.du/dx=x/x=1. Substituting these into the 'by p...

To solve this question we will use the chain rule, as we can see that we have one function being applied to another, i.e sin is being applied to cos(x).

This means we are able to replace the origin...