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Chapter 6: Problem 13

If \(y=\frac{\sec x+\tan x-1}{\sec x-\tan x+1}\), find \(\frac{d y}{d x}\) at \(x=0\)

Short Answer

Expert verified
The derivative of the function at \(x=0\) is 2.

Step by step solution

01

Apply the Quotient Rule

To find \(\frac{d y}{d x}\) for the function \(\frac{u}{v}\), the Quotient Rule states that \(\frac{d y}{d x} = \frac{v*(du/dx) - u*(dv/dx)}{v^2}\). Here \(u=\sec x+\tan x-1\) and \(v=\sec x-\tan x+1\). First find the derivatives of \(u\) and \(v\) which are \(du/dx= \sec x \tan x + \sec^2 x\) and \(dv/dx= \sec x \tan x - \sec^2 x\). Now apply the Quotient Rule.
02

Simplify the Derivative Expressions

We now need to plug the expressions for \(u\), \(v\), \(du/dx\), and \(dv/dx\) into the Quotient Rule formula. Doing so, we get: \( \frac{d y}{d x} = \frac{(\sec x-\tan x+1)(\sec x \tan x + \sec^2 x) - (\sec x+\tan x-1)(\sec x \tan x - \sec^2 x)}{(\sec x-\tan x+1)^2}\). We can simplify this to \(\frac{d y}{d x} = 2\sec^2 x\).
03

Evaluate the derivative at x=0

With the derivative function, now substitute \(x=0\) into the function for \(\frac{d y}{d x}\). Thus, we have \(\left(\frac{d y}{d x}\right)_{x=0} = 2\sec^2 0 = 2\), as \(\sec 0 = 1\).

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Most popular questions from this chapter

If \(y=e^{x}(\sin x+\cos x)\), prove that \(y_{2}-2 y_{1}+2 y=0\)

If \(y=\cos ^{-1}\left(\sqrt{\frac{\cos 3 x}{\cos ^{3} x}}\right)\), prove that \(\frac{d y}{d x}=\sqrt{\frac{6}{\cos 2 x+\cos 4 x}}\)

If \(f(x)=x^{3}+2 x^{2}+3 x+4\) and \(g(x)\) is the inverse of \(f(x)\), find \(g^{\prime}(4)\).

If \(y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x}}}+\ldots\) to \(\infty\), prove that \(\frac{d y}{d x}=\frac{\cos x}{2 y-1}\)

If \(y=e^{\tan ^{-1}} x\), show that $$ \left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0 $$
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