Question #1506a

Question

Question #1506a

2 Answers

Sean

Nov 21, 2017

#

Answer link

Impact of this question

1074 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-21T23:33:51

$\frac{d y}{d x} = \frac{1}{\sqrt{1 + x^{2}}}$ $dy/dx = 1/sqrt(1+x^2)$

Explanation:

You can simply use the chain rule:

$\frac{d}{d x} ln (x + \sqrt{1 + x^{2}}) = (\frac{1}{x + \sqrt{1 + x^{2}}}) \frac{d}{d x} (x + \sqrt{1 + x^{2}})$ $d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) )d/dx(x+sqrt(1+x^2))$

$\frac{d}{d x} ln (x + \sqrt{1 + x^{2}}) = (\frac{1}{x + \sqrt{1 + x^{2}}}) (1 + \frac{x}{\sqrt{1 + x^{2}}})$ $d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (1+x/sqrt(1+x^2))$

$\frac{d}{d x} ln (x + \sqrt{1 + x^{2}}) = (\frac{1}{x + \sqrt{1 + x^{2}}}) \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ $d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (x+sqrt(1+x^2))/sqrt(1+x^2)$

$\frac{d}{d x} ln (x + \sqrt{1 + x^{2}}) = \frac{1}{\sqrt{1 + x^{2}}}$ $d/dx ln(x+sqrt(1+x^2)) = 1/sqrt(1+x^2)$

or you can note that:

$f (x) = ln (x + \sqrt{1 + x^{2}})$ $f(x) = ln(x+sqrt(1+x^2))$

is the logarithmic form of the inverse hyperbolic sine function so that:

$y = ln (x + \sqrt{1 + x^{2}}) \Leftrightarrow sinh y = x$ $y= ln(x+sqrt(1+x^2)) <=> sinhy = x$

differentiating the second expression implicitly:

$\frac{d}{d x} (sinh y) = 1$ $d/dx (sinhy) = 1$

$cosh y \frac{d y}{d x} = 1$ $coshy dy/dx = 1$

$\frac{d y}{d x} = \frac{1}{cosh y}$ $dy/dx = 1/coshy$

and from the identity:

${cosh}^{2} y - {sinh}^{2} y = 1$ $cosh^2y-sinh^2y = 1$

we then get:

$cosh y = \sqrt{1 + {sinh}^{2} y} = \sqrt{1 + x^{2}}$ $coshy = sqrt(1+sinh^2y) = sqrt(1+x^2)$

and:

$\frac{d y}{d x} = \frac{1}{\sqrt{1 + x^{2}}}$ $dy/dx = 1/sqrt(1+x^2)$

Science

Math

Humanities

... and beyond

Question #1506a

2 Answers

Explanation:

Related questions

Impact of this question