# Thread: Implicit Differentiation When the Tangent is Defined

1. ## Implicit Differentiation When the Tangent is Defined

Find the slope of tangent to x^2y^3=x^4-y^4-1 at x=1 when the tangent is defined.

Slope of tangent line m=?

I having trouble finding a tangent that is defined as I'm not sure how to make it so the function is defined at the point where x=1.

Thanks, DLL.

2. Originally Posted by Daddy_Long_Legs
Find the slope of tangent to x^2y^3=x^4-y^4-1 at x=1 when the tangent is defined.

Slope of tangent line m=?

I having trouble finding a tangent that is defined as I'm not sure how to make it so the function is defined at the point where x=1.

Thanks, DLL.
Start by using implicit differentiation (as your post title says) to get an expression for dy/dx. Can you do this? If not, where do you get stuck?

3. $x^{2}y^{3} = x^{4}-y^{4} -1$

Differentiating implicitly and using the product rule you get

$2x*y^{3} + x^{2}*3y^{2} \frac {dy}{dx} = 4x^3 = 4y^{3}\frac{dy}{dx}$

solve for $\frac {dy}{dx}$

4. mr fantastic,

I can use implicit differentiation to find dy/dx. The problem then comes when I try to find the gradient or m, as the the value x=1 makes y=0 and this therefore makes the function undefined when solving for m, but I'm not sure how to find the gradient when it's defined?

Thanks for the quick relpy, DLL.

5. Random Variable,

=4x^3-2xy/3y^2x^2+4y^3

Now when I try to solve this using the values x=1 and y=0 (y=0 is from solving y using the value of x=1) the gradient or m become 4/0 which is undefined.

The trouble now comes with trying to define this function as I'm not sure how to?

Thanks for the reply, DDL.

6. Originally Posted by Daddy_Long_Legs
mr fantastic,

I can use implicit differentiation to find dy/dx. The problem then comes when I try to find the gradient or m, as the the value x=1 makes y=0 and this therefore makes the function undefined when solving for m, but I'm not sure how to find the gradient when it's defined?

Thanks for the quick relpy, DLL.
There are two values of y when x = 1. Please post your working for finding for y.

7. mr fantastic,

I have done this using the value of x=1 and substituting it into the expression.

(1)^2y^3=(1)^4-y^4-1
y^3=1-y^4-1
y^3-y^4=0

There I have y=0 which make the function undefined. I more than likely have make an error here, but ican't see where .

Thanks, DLL.

8. Originally Posted by Daddy_Long_Legs
mr fantastic,

I have done this using the value of x=1 and substituting it into the expression.

(1)^2y^3=(1)^4-y^4-1
y^3=1-y^4-1 Mr F says: Correct.
y^3-y^4=0 Mr F says: Wrong. It's y^3 + y^4 = 0

There I have y=0 which make the function undefined. I more than likely have make an error here, but ican't see where .

Thanks, DLL.
$y^3 + y^4 = 0 \Rightarrow y^3 (1 + y) = 0 \Rightarrow y = 0, \, -1$.

9. mr fantastic,

Sorry I did mean y^3+y^4=0.

Anyway, I didn't realise I could factorise it after this step.

Thanks a lot for that, DLL.