Slippery slopes... another derivative
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You might find some of the ideas in Slippery slopes helpful when thinking about this problem.

Here are the graphs of $y=\dfrac{1}{x}$ and $y=\dfrac{1}{x+1}$ with tangents drawn at points A and B.

What happens when you move the point A? (Be careful when the $x$ -coordinate of A gets close to $-1$ .)

How could you work out the gradient of $y=\dfrac{1}{x+1}$ at the point $(2,\tfrac{1}{3})$ ?
How can you differentiate $\dfrac{1}{x+1}$ ?
How could you adapt these approaches to work out the derivative of $\dfrac{1}{2x+1}$ ?

The graph of $y=\dfrac{1}{x+1}$ is a translation of the graph of $y=\dfrac{1}{x}$ .

Can you generalise ideas from part 1 to find the derivative of $\dfrac{1}{x+1}$ at a general point $(x,y)$ ?

We already know how to differentiate constant multiples of powers of $x$ , so we know how to differentiate $\dfrac{1}{2x}$ .

Can you suggest other examples where you could use a similar approach to find derivatives?

In this problem we have only composed the function $\dfrac{1}{x}$ with functions of the form $ax+b$ . Could you apply any of these ideas to polynomial, rational or trigonometric functions?

Chain Rule & Integration by Substitution

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Slippery slopes... another derivative
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