How do you find the derivative of $y = \frac{sec x}{1 + tan x}$ $y = secx/(1 + tanx)$ ?

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How do you find the derivative of $y = \frac{sec x}{1 + tan x}$ $y = secx/(1 + tanx)$ ?

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Answer 1 · 2016-12-01T02:58:01

$\frac{d y}{d x} = \frac{sin x - cos x}{{(sin x + cos x)}^{2}}$ $dy/dx = (sinx - cosx)/(sinx + cosx)^2$

Explanation:

First of all, I'm assuming the function is $y = \frac{sec x}{1 + tan x}$ $y = secx/(1 + tanx)$ ? Call your function $f (x)$ $f(x)$ .

$f (x) = \frac{sec x}{1 + tan x}$ $f(x) = secx/(1 + tanx)$

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

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

$f (x) = \frac{1}{cos x} \times \frac{cos x}{cos x + sin x}$ $f(x) = 1/cosx xx cosx/(cosx + sinx)$

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

$f (x) = {(cos x + sin x)}^{- 1}$ $f(x) = (cosx + sinx)^-1$

We let $y = u^{- 1}$ $y = u^-1$ and $u = cos x + sin x$ $u = cosx + sinx$ . Then $\frac{d y}{d u} = - 1 u^{- 2}$ $dy/(du) = -1u^(-2)$ and $\frac{d u}{d x} = - sin x + cos x$ $(du)/dx = -sinx + cosx$ .

The chain rule states that $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $color(red)(dy/dx = dy/(du) xx (du)/dx$ .

$\frac{d y}{d x} = - \frac{1}{u^{2}} \times - sin x + cos x$ $dy/dx = -1/u^2 xx -sinx + cosx$

$\frac{d y}{d x} = \frac{sin x - cos x}{{(sin x + cos x)}^{2}}$ $dy/dx = (sinx - cosx)/(sinx + cosx)^2$

Hopefully this helps!

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How do you find the derivative of $y = \frac{sec x}{1 + tan x}$ $y = secx/(1 + tanx)$ ?

1 Answer

Explanation:

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How do you find the derivative of $y = \frac{sec x}{1 + tan x}$ $y = secx/(1 + tanx)$ ?