Question

Implicit differentiation calculus,

can you help me solve the first one please.

https://www.dropbox.com/s/93h9x2h270cdhk8/Screenshot%202015-12-07%2016.14.45.png?dl=0

Answer 1

Treat y as a function of x and use the chain rule
x^2 + y^2 = 100
2x + 2y y' = 0

solve for y'

Answer 2

-x/y

Answer 3

No 4 conatains 3xy^2
- here you use the product rule/chain rule.

Answer 4

correct

Answer 5

why if I first solve for y and differentiate it gives different result ? @welshfella

Answer 6

- that way you get the derivative in terms of x . It will have the same value as the derivative in terms of x and y.

Answer 7

can you please explain to me what you just said ?

Answer 8

am I not looking for dy/dx in both cases ?

Answer 9

the derivative will be a different expression but there will only be x in it - no y's.
yes

Answer 10

but they will be the same thing both of them, yes ?

Answer 11

they will both be the slope of a tangent to the curve at any point

Answer 12

same tangent , right?

Answer 13

ah i see

Answer 14

Its not always possible to find y in terms of x . That;s when implicit differentation is useful.

Answer 15

so implicit differentiation helps you find it in terms of x y

Answer 16

yes

Answer 17

thanks

Answer 18

yw

Answer 19

so just to sum up, the anwer to #4 is just -x/y ? no further steps ?

Answer 20

thats correct

Answer 21

ok

Answer 22

**why if I first solve for y and differentiate it gives different result ?***

You should get the same answer (though sometimes solving for y is not very practical)
For example , Question 3-12, 3.
\[ x^2+y^2= 100 \]
by implicit differentiation you get
\[ y' = -\frac{x}{y}\]

if you solve for y:
\[ y= \sqrt{100-x^2} \]
and
\[ y'= \frac{1}{2} \frac{-2x}{\sqrt{100-x^2}} \\
y'= - \frac{x}{\sqrt{100-x^2}}
\]
if you replace the denominator with the equivalent y you get
\[ y'= -\frac{x}{y}\]

Answer 23

how would you replace the denominator with the equivalent y ? @phi

Answer 24

when you solved for y, you got
\[ y= \sqrt{100-x^2} \]
notice the right-hand side is the denominator in y'
so you can put in y instead of \( \sqrt{100-x^2} \)

Answer 25

we should prob use \(y= \pm\sqrt{100-x^2}\)

then \(\displaystyle y'=\pm\frac{1}{2}\frac{-2x}{\sqrt{100-x^2}}\)

simplify...

\[y'=\pm\frac{-x}{\sqrt{100-x^2}}=\frac{-x}{\pm\sqrt{100-x^2}}\]

replace \(y\)

\[y'=\frac{-x}{y}\]