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One of the best ways to view dy/dx is as a fraction. When we have y=f(g(x)), we need to make a substitution u=g(x) to find dy/dx. This leaves us y=f(u) and u=g(x). Differentiating said terms leaves us wit...

First, we find the form of the two fractions we're going to get. As one denominator has a power of 2, and the other a power of 1, our answer will be of the form: [(Ax+B)/(x²+1)] + [C/(x+2)]. If...

x = log5/log2

Stationary points occurs when the gradient of the graph is equal to 0, i.e. dy/dx = 0. Differentiate y with respect to x to get dy/dx = 4x + 4.So making 4x + 4 = 0 gives x = -1. Substituting this into the...

When you differentiate the function that gives the gradient. Therefore differentiate to get dy/dx= 15x^2 + 3. Make this equal to 0 to get 15x^2 + 3=0 and rearrange to get the value of x as required so 15x...