Find dy/dx for Implicit Differentiation of cos(xy^2)-(x^3+y)/(x+1)=sec(x)sin(y)

In summary, implicit differentiation is needed to find dy/dx given that: cos(xy^2) - (x^3+y) / (x+1) = sec(x)sin(y).
  • #1
MrRottenTreats
18
0
assistance asap is needed !

Using implicit differentiation, find dy/dx given that:
cos(xy^2) - (x^3+y) / (x+1) = sec(x)sin(y)

i am horrid at these, i came up with a few lines, please check them

-sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sin(y)(dy/dx)

after this line i have no idea what else i can do with this, it looks just looks like a huge mess to me.
 
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  • #2
You seem to have left out the y (in the numerator) while using the quotient rule, and you need a sec(x) (not sin(y) - second term RHS).

Try to bring the dy/dx s to one side and see if the rest of the expressions simplify, somehow.
 
  • #3
so it would look this so:

-sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)(dy/dx)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sec(y)(dy/dx)


so many dy/dx 's should i still try and bring them all on one side?
 
  • #4
MrRottenTreats said:
so it would look this so:

-sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)(dy/dx)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sec(y)(dy/dx)

You missed the dy/dx again...

{(3x^2 + dy/dx)(x+1) - (x^3+y)} / (x+1)^2

so many dy/dx 's should i still try and bring them all on one side?

Yes, if the question requires it. I think many terms will cancel if you expand the expression on both sides.
 
  • #5
ok i get that now, but when i try to isolate the dy/dx the big bracket is too hard to get the (dy/dx) out of..

this is what i got when i expanded it.

[{(3x^3)+(3x^2)+(x)(dy/dx)+(dy/dx)-(x^3)-(y)} / (x+1)^2 ]

then only 2 terms in then can be groups the 3x^3 - x^3 and i don't know where to do for the rest of it..
 

1. What is implicit differentiation?

Implicit differentiation is a method used to find the derivative of an equation that is not explicitly written in terms of the independent variable. It involves treating the dependent variable as a function of the independent variable and using the chain rule to find the derivative.

2. How do you use implicit differentiation to find dy/dx?

To find dy/dx using implicit differentiation, differentiate both sides of the equation with respect to x. Use the chain rule to differentiate any terms that contain the dependent variable, and then solve for dy/dx.

3. Can you explain the steps for finding dy/dx using implicit differentiation?

Step 1: Identify the terms that contain both x and y.Step 2: Apply the chain rule to differentiate these terms.Step 3: Isolate the term containing dy/dx on one side of the equation.Step 4: Simplify and solve for dy/dx.

4. How do you handle trigonometric functions in implicit differentiation?

When differentiating trigonometric functions in implicit differentiation, you can use the chain rule and the derivative rules for trigonometric functions. For example, the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x).

5. What is the derivative of a constant in implicit differentiation?

The derivative of a constant in implicit differentiation is 0. This is because a constant is not a function of the independent variable, so it does not change as the independent variable changes.

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