What is the equation of the tangent line of $f (x) = cot x \cdot sec x$ $f(x)=cotx*secx$ at $x = \frac{π}{4}$ $x=pi/4$ ?

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What is the equation of the tangent line of $f (x) = cot x \cdot sec x$ $f(x)=cotx*secx$ at $x = \frac{π}{4}$ $x=pi/4$ ?

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Answer 1 · 2015-12-22T16:14:01

$y - \sqrt{2} = \sqrt{2} (x - \frac{π}{4})$ $y-sqrt2=sqrt2(x-pi/4)$

Explanation:

First, simplify $f (x)$ $f(x)$ .

$f (x) = \frac{cos x}{sin x} (\frac{1}{cos x}) = \frac{1}{sin x} = csc x$ $f(x)=cosx/sinx(1/cosx)=1/sinx=cscx$

Thus, $\frac{d}{d x} [csc x] = - csc x cot x$ $d/dx[cscx]=-cscxcotx$

This is identity you should know. It can be proven through the qoutient rule $\frac{d}{d x} [\frac{1}{sin x}]$ $d/dx[1/sinx]$ or the chain rule ${(sin x)}^{- 1}$ $(sinx)^-1$ .

Since $f'(x)=-cscxcotx$ , we know that:

$f'(pi/4)=-cot(pi/4)csc(pi/4)=-1(sqrt2)=-sqrt2$

We also know that

$f(pi/4)=sqrt2$

Thus, the tangent line passes through the point $(pi/4,sqrt2)$ and has a slope of $-sqrt2$ .

Write this in point-slope form:

$y-sqrt2=-sqrt2(x-pi/4)$

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What is the equation of the tangent line of $f (x) = cot x \cdot sec x$ $f(x)=cotx*secx$ at $x = \frac{π}{4}$ $x=pi/4$ ?

1 Answer

Explanation:

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What is the equation of the tangent line of f(x)=cotx*secxf(x)=cotx⋅secxf(x)=cotx*secx at x=pi/4x=π4x=pi/4?

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What is the equation of the tangent line of $f (x) = cot x \cdot sec x$ $f(x)=cotx*secx$ at $x = \frac{π}{4}$ $x=pi/4$ ?