Question

Find dy/dx by implicit differentiation. y^2+xy=x^2+1

Answer 1

Please you have to compute the first derivative of both sides of your equation, using the chain-rule

Answer 2

I don't understand the chain rule.

Answer 3

How would I derive xy

Answer 4

for example, I apply the cahin-rule as follows:
\[\frac{d}{{dx}}\left( {xy} \right) = y + xy'\]

Answer 5

okay.
would the 2nd part be just x?

Answer 6

no, since you have y=y(x) and please keep in mind you are computing the first derivative with respect to the x-variable

Answer 7

hint:
\[\frac{d}{{dx}}\left( {{y^2}} \right) = 2yy'\]

Answer 8

Thank you for being kind to me, I'm so hopelessly lost.

Answer 9

that is the chain-rule

Answer 10

My textbook doesn't even mention that.
So it is second derivative?

Answer 11

no it is the first derivative, it is the first deivative of a function of function. namely:
\[\frac{{d{y^2}}}{{dx}} = \frac{{d{y^2}}}{{dy}} \cdot \frac{{dy}}{{dx}} = 2y \cdot y'\]

Answer 12

please you have to search for first derivative of function of function, or first derivative of composite functions, on your textbook

Answer 13

according to my textbook a first derivative function is: \[y ^{1}=\frac{ dy }{dx }=f \prime \left( x \right)\]

Answer 14

how would I incorporate that into the question?

Answer 15

yes! that is the fist derivative of y=f(x), nevertheless kep in mind that yo have this function:
\[g\left( x \right) = y\left( x \right) \cdot y\left( x \right)\]
Next I apply the Leibniz rule, and I get:
\[\frac{d}{{dx}}g\left( x \right) = \frac{{dy}}{{dx}} \cdot y + y \cdot \frac{{dy}}{{dx}} = 2y \cdot \frac{{dy}}{{dx}}\]

Answer 16

oops.. nevertheless keep in mind...

Answer 17

I do not believe we covered the Leibniz rule

Answer 18

Where does the 2y come from?

Answer 19

please note that sometime the "Leibniz rule" is called "product rule"

Answer 20

is that the y=uv?

Answer 21

there is 2y since I have added similar terms, namely:
\[\frac{{dy}}{{dx}} \cdot y + y \cdot \frac{{dy}}{{dx}} = y \cdot \frac{{dy}}{{dx}} + y \cdot \frac{{dy}}{{dx}}\]
since:
\[\frac{{dy}}{{dx}} \cdot y = y \cdot \frac{{dy}}{{dx}}\]

Answer 22

yes, it is the same rule that we have to apply in order to find the derivative of y=u*v

Answer 23

okay that makes sense. Since you've applied both rules to the left side of the equation. What do I do with the right side?

Answer 24

the right side is very easy, please you have to apply this identity:
\[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\]

Answer 25

furthermore, please keep in mind that, we have:
\[\frac{d}{{dx}}1 = 0\]

Answer 26

okay so for the end of this problem do I come up with a number for an answer or another equation?

Answer 27

Because we aren't using real numbers necessarily, just applying rules. Do these rules satisfy the equation?

Answer 28

yes! all theorems of differential calculus can be applied to all problems which involve real functions of real variable, namely y=f(x)

Answer 29

of course we can apply those theorems also in order to solve problems with implicit functions

Answer 30

and the equation I have is implicit.

Answer 31

yes!

Answer 32

Okay so first I have to solve for the left side by using the chain rule then the product rule? And for the right side you said to use the power rule?

Answer 33

yes!

Answer 34

are there any other rules I need to apply?

Answer 35

another rule that you can have to apply is this:
\[\frac{d}{{dx}}\left( {f + g} \right) = \frac{d}{{dx}}f + \frac{d}{{dx}}g\]

Answer 36

but I do not have g? How do I apply this rule?

Answer 37

for example at the left side you have these function:
\[\begin{gathered}
f = {y^2} \hfill \\
g = xy \hfill \\
\end{gathered} \]
please note that the symbols f and g are used in a general sense here

Answer 38

OH okay! That clears it up

Answer 39

if you like I can rewrite the above identity as follows:
\[\frac{d}{{dx}}\left( {P + Q} \right) = \frac{d}{{dx}}P + \frac{d}{{dx}}Q\]

Answer 40

and:
\[\begin{gathered}
P\left( x \right) = {y^2} \hfill \\
Q\left( x \right) = xy \hfill \\
\end{gathered} \]

Answer 41

\[\frac{ d }{ dx }(y^2+xy)=\frac{ d }{ dx }(y^2)+\frac{ d }{ dx }(xy)\]
is that what it should look like?

Answer 42

ok!

Answer 43

how would I go about to simplify that further? Is this the part where I apply the chain rule?

Answer 44

yes!

Answer 45

okay now I've solved it. Thank you so much for your patience!

Answer 46

Thank you!

Answer 47

what is the volume in cubic inches of a cube whose total surface area is 96 square inches?

Answer 48

can you help me out about the question?

Answer 49

you should make a new post for your new question.
but to answer it, notice that a cube has 6 (all equal ) sides, so one side has area 96/6
= 16 square inches.
now you can figure out the length of 1 side of the square.
the volume will be the side to the 3rd power