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Let y = tan⁻¹2x/(1 - x²) for -1 < x < 1…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Let $y = \tan^{-1}\dfrac{2x}{1 - x^2}$ for $-1 < x < 1$ . Put $x = \tan\theta$ to simplify $y$ . Then find $\dfrac{dy}{dx}$ at $x = 0.5$ and at $x = 0$ .

Need a reminder? The key ideaShow

$(e^x)' = e^x$ , $(a^x)' = a^x\ln a$ , $(\ln x)' = \dfrac{1}{x}$ , $(\log_a x)' = \dfrac{1}{x\ln a}$ , $(\sin^{-1}x)' = \dfrac{1}{\sqrt{1 - x^2}}$ , $(\tan^{-1}x)' = \dfrac{1}{1 + x^2}$ . Combine each with the chain rule.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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← Let y = ln(x² + 4). Find dy/dx at x = 2.Let y = log₁₀(x² + 1). Find dy/dx at x = 3, to four decimal places. Use ln 10 ≈ 2.302585. →

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