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Calculus, EXT1 EQ-Bank 16

Find the equation of the tangent to the curve  \(y=\tan ^{-1}\left(x^2\right)\)  at the point on the curve where  \(x=1\).   (3 marks)

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\(y=x-1+\dfrac{\pi}{4}\)

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\(y=\tan ^{-1}\left(x^2\right) \ \Rightarrow \ \dfrac{dy}{dx}=\dfrac{1}{1+\left(x^2\right)^2} \times 2 x=\dfrac{2 x}{1+x^4}\)

\(\text{At} \ \ x=1:\)

\(y=\tan ^{-1}\left(1^2\right)=\dfrac{\pi}{4}\)

\(\dfrac{dy}{dx}=\dfrac{2}{1+1^4}=1\)
 

\(\text{Find equation of tangent,} \ \ m=1 \ \ \text {through}\ \left(1, \dfrac{\pi}{4}\right):\)

\(y-\dfrac{\pi}{4}\) \(=1(x-1)\)
\(y\) \(=x-1+\dfrac{\pi}{4}\)

Calculus, EXT1 EQ-Bank 19

The function  \(y=f(x)\)  has an inverse function  \(y=f^{-1}(x)\).

The tangent to  \(y=f(x)\) at the point \((2,3)\), \(\ell\), has a gradient of 1.

Show that the tangent to  \(y=f^{-1}(x)\)  at the point \((3,2)\) is parallel to \(\ell\).   (3 marks)

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\(\text{See Worked Solution}\)

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\(\text{Since tangent to}\  f(x) \ \text {touches at}\ (2,3):\)

\(f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2\)
 

\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)

\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)

\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
 

\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)

\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)

\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).

Calculus, EXT1 EQ-Bank 32

The polynomial  \(h(x)=x^3+2x+1\)  passes through the point \((1,4)\).

Find the gradient of the tangent to  \(f(x)=x h^{-1}(x)\)  at the point where \(x=4\).   (3 marks)

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\(\dfrac{9}{5}\)

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\(h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1\)

\(f(x)=x h^{-1}(x)\)

\(\text{Using the product rule:}\)

\(f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)\)
 

\(\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):\)

\(\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)\)

\(\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}\)

\(\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}\)
 

\(f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}\)

\(f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}\)
 

\(h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2\)

\(f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}\)

\(\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}\)

Calculus, EXT1 C2 EQ-Bank 30

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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\(-\dfrac{\pi^2}{32}\)

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\(y=\dfrac{\cos2x}{x^2}\)

\(\text{Inverse: Swap} \ \ x \leftrightarrow y\)

\(x=\dfrac{\cos 2 y}{y^2}\)

\(u=\cos 2 y \ \ \quad \quad \quad v=y^2\)

\(v^{\prime}=-2 \sin 2 y \ \ \quad v^{\prime}=2 y\)

\(\dfrac{dx}{dy}=\dfrac{-2 y^2 \sin 2 y-2 y\, \cos 2 y}{y^4}\)

\(\dfrac{dy}{dx}=\dfrac{y^4}{-2 y^2 \sin 2 y-2 y\, \cos 2 y}\)
 

\(\text{At}\ \ y=\dfrac{\pi}{4}:\)

\(\dfrac{d y}{d x}\) \(=\dfrac{\left(\dfrac{\pi}{4}\right)^4}{-2\left(\dfrac{\pi}{4}\right)^2 \sin \left(\dfrac{\pi}{2}\right)-2\left(\dfrac{\pi}{4}\right) \cos \left(\dfrac{\pi}{2}\right)}\)
  \(=-\dfrac{\pi^4}{256} \times \dfrac{8}{\pi^2}\)
  \(=-\dfrac{\pi^2}{32}\)

Calculus, EXT1 C2 2022 HSC 12c

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`

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`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`