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Find \(d y / d x\) by implicit differentiation. $$ x=\sec \frac{1}{y} $$

Short Answer

Expert verified
\( dy/dx = -y^2\sec \left( \frac{1}{y} \right) \cdot \tan \left( \frac{1}{y} \right) \)

Step by step solution

01

Differentiate both sides with respect to x

On the left side, the derivative of x with respect to x is just 1. On the right side, the derivative of \( \sec \frac{1}{y} \) with respect to x is more complex. Recognise that this function can be considered a 'function within a function', that's why we use the chain rule. So firstly, differentiate \( \sec(u) \) where \(u = \frac{1}{y}\) to get \( \sec(u)\tan(u) \). Secondly, differentiate \( u = \frac{1}{y} \) to get \( -\frac{1}{y^2} \). Multiply these two results by the chain rule. So we have: \[ 1 = \sec \left( \frac{1}{y} \right) \cdot \tan \left( \frac{1}{y} \right) \cdot -\frac{1}{y^{2}} \]
02

Solve for dy/dx

Now, rearrange the equation to solve for dy/dx. To isolate dy/dx on one side, multiply the entire equation by -y^2. \[ -y^2 = \sec \left( \frac{1}{y} \right) \cdot \tan \left( \frac{1}{y} \right) \] There you have it! You've found \( dy/dx \) as a function of x and y, using implicit differentiation.

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