1. ## Implicit differentatation?

I am having difficulty differentiating the following questions

1. x^(2/3)+y(2/3)=5 This is what I did

(2/3)x^(-1/3)+(2/3)y^(-1/3) I am also supposed to plot the point(8,1) to find slope

2. tan(x+y)=x This is what I did

sec^2(x+y)(1+1)=1 ? (is this correct?)

3. y=sin(xy)

1=cos(xy)(1y+1x) is this correct?

2. ## Re: Implicit differentatation?

while differentiating variable y you have to write dy/dx
as in question 1 (2/3)x^(-1/3) + (2/3)y^(-1/3)dy/dx = 0

3. ## Re: Implicit differentatation?

Originally Posted by waqarhaider
while differentiating variable y you have to write dy/dx
as in question 1 (2/3)x^(-1/3) + (2/3)y^(-1/3)dy/dx = 0
From here, plug in x = 8 and y = 1 to solve for (dy/dx).

For 3. NO.
The derivative of y is NOT ONE!!
The derivative of y is
dy/dx, or y'.

You should have y' = cos(xy)*(xy)' -->
y' = cos(xy)*(xy' + yx'), but x' IS ONE, so we can write this as
y' = cos(xy)*(xy' + y), or
dy/dx = cos(xy)*(x*dy/dx + y)

4. ## Re: Implicit differentatation?

So would the answer to my second question be

sec^2(x+y)(1+dy/dx)=1

Also on a similar query what is meant by d^2y/dx^2 in terms of x and y.

For a function such as x^2+xy=5

Does that mean to differentiate two times

5. ## Re: Implicit differentatation?

yes, the derivative of the derivative ...

$\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

6. ## Re: Implicit differentatation?

tan(x+y)=x

sec^2(x+y)(1+dy/dx)=1 I am wondering did get the derivative correctly?

7. ## Re: Implicit differentatation?

Originally Posted by homeylova223
tan(x+y)=x

sec^2(x+y)(1+dy/dx)=1 I am wondering did get the derivative correctly?
so far so good, but get dy/dx by its lonesome ...