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Given \(y=\dfrac{\cos 2 x}{x^2}\), find the gradient of the tangent of its inverse function at \(\left(\dfrac{8 \sqrt{2}}{\pi^2}, \dfrac{\pi}{4}\right)\). (3 marks)
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\(-\dfrac{\pi^2}{32}\)
\(y=\dfrac{\cos2x}{x^2}\)
\(\text{Inverse: Swap} \ \ x \leftrightarrow y\)
\(x=\dfrac{\cos 2 y}{y^2}\)
\(u=\cos 2 y \ \ \quad \quad \quad v=y^2\)
\(v^{\prime}=-2 \sin 2 y \ \ \quad v^{\prime}=2 y\)
\(\dfrac{dx}{dy}=\dfrac{-2 y^2 \sin 2 y-2 y\, \cos 2 y}{y^4}\)
\(\dfrac{dy}{dx}=\dfrac{y^4}{-2 y^2 \sin 2 y-2 y\, \cos 2 y}\)
\(\text{At}\ \ y=\dfrac{\pi}{4}:\)
| \(\dfrac{d y}{d x}\) | \(=\dfrac{\left(\dfrac{\pi}{4}\right)^4}{-2\left(\dfrac{\pi}{4}\right)^2 \sin \left(\dfrac{\pi}{2}\right)-2\left(\dfrac{\pi}{4}\right) \cos \left(\dfrac{\pi}{2}\right)}\) |
| \(=-\dfrac{\pi^4}{256} \times \dfrac{8}{\pi^2}\) | |
| \(=-\dfrac{\pi^2}{32}\) |
Find the equation of the tangent to the curve `y=x text{arctan}(x)` at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form `y=mx+c` (3 marks)
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`y=((2+pi)/4)x-1/2`
| `y` | `=xtan^(-1)(x)` | |
| `dy/dx` | `=x xx 1/(1+x^2)+tan^(-1)(x)` |
`text{When}\ \ x=1:`
`dy/dx=1/2+tan^(-1)(1)=1/2+pi/4=(2+pi)/4`
`text{Equation of tangent}\ \ m=(2+pi)/4,\ text{through}\ \ (1,(pi)/(4)):`
| `y-pi/4` | `=(2+pi)/4 (x-1)` | |
| `y` | `=((2+pi)/4)x-(2+pi)/4+pi/4` | |
| `y` | `=((2+pi)/4)x-1/2` |