Question

By first differentiating (√(1-x))/(√(1+x)) obtain an expression dy/dx in terms of x. Hence show that the gradient of the normal to the curve at point (x,y) is (1+x)(√(1-x^2))

Answer 1

\[\frac{ \sqrt{1-x} }{ \sqrt{1+x} }\]

Answer 2

gradient of normal = -1/(gradient of tangent)

= \(-\dfrac{1}{\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)'}\)

Answer 3

I found dy/dx i.e \[\frac{ -2 }{ (1+x^{2}) }\]
How do I show that the gradient of the normal is equal to \[(1+x)\sqrt{1-x ^{2}}\]

Answer 4

dy/dx looks wrong.
im(wolfram) getting : dy/dx = http://www.wolframalpha.com/input/?i=%28%5Cfrac%7B+%5Csqrt%7B1-x%7D+%7D%7B+%5Csqrt%7B1%2Bx%7D+%7D%29%27

Answer 5

taking the reciprocal of above and negating it gives the expression you looking for

Answer 6

No, actually the question says just to derivate (1+x)/(1-x) and then use the answer to obtain the gradient of the normal.

Answer 7

yes, so ?

Answer 8

The marking scheme I have says to use a chain rule. I don't know how to do that.

Answer 9

im not getting u friend :/

Answer 10

Okay, I'll post the screenshot.

Answer 11

okay

Answer 12

how come the radical sign disappeared ?

Answer 13

doesnt that look like a clear typo ?

Answer 14

I don't know. This is what the marking scheme says.

Answer 15

Gotcha :)

Answer 16

you're right, they want us to use chain rule lol

Answer 17

\(f(g(x)) = \sqrt{\dfrac{1-x}{1+x}}\)

Answer 18

\(g(x) = \dfrac{1-x}{1+x}\)

\(f(x) = \sqrt{x}\)

Answer 19

\([f(g(x))]' = f'(g(x)) \times g'(x)\)

Answer 20

plug and chug ?

Answer 21

\(g(x) = \dfrac{1-x}{1+x} \implies g'(x) = \dfrac{-2}{(1+x)^2}\)

Answer 22

Okay. I did that too.
f'(x)= \[1/2\sqrt{x}\]

Answer 23

\(f(g(x)) = \sqrt{\dfrac{1-x}{1+x}} \)

\(\implies [f(g(x))]' = \dfrac{1}{2\sqrt{\dfrac{1-x}{1+x}}} \times \left(\dfrac{-2}{(1+x)^2}\right) \)

Answer 24

simplify

Answer 25

I am still not getting that expression.
All I got was:
\[\sqrt{1-x} (1+x)^{3/2}\]

Answer 26

^This was after taking the negative inverse of the gradient.

Answer 27

Lol you're done then

Answer 28

just rearrange that expression

Answer 29

\(\sqrt{1-x} (1+x)^{3/2} = \sqrt{1-x} (\sqrt{1+x})^{3} \\ =\sqrt{1-x}\sqrt{1-x} (\sqrt{1+x})^{2} = \sqrt{1-x^2}(1+x)\)

Answer 30

O Lord, this is such a bad question. Too much simplification. How could have anyone done it during an exam with under 5 minutes per question? -_-

Answer 31

@ganeshie8 : Thank you. Much appreciated! :)

Answer 32

hahah its more of algebra... not calculus lol

Answer 33

u wlc :)