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In order to find the gradient of the curve at (1,1), we must first differentiate the equation of the curve. To do this, multiply the coefficient of x by the power of that same x. Then subtract one from th...

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The tangent is the straight line passing through x=1, touching the curve only at that point. For x=1, y=(4+1)^3=125 Using the chain rule we obtain dy/dx = 38x(4x^2+1)^2. To then get the gradient ...

I often find it difficult to remember all the different identities, so what is useful is instead to just remember the familiar identity sin^2(x) + cos^2(x) = 1 that we have come across many times, and div...

Recall the product rule for differentiation. If y=uv, where u and v are functions defined by functions of x, then we can take the derivative of y as: y'=u'v+v'u () (where ' denotes the derivative) ...