It's correct with the added parentheses.dylanhouse said:Homework Statement
I need to use implicit differentiation to find the derivative of y=sin(x+y).
Homework Equations
The Attempt at a Solution
This is what I did:
y=sin(x+y)
y'=(sin(x+y))'
y'=(1+y')(cos(x+y)) (by the chain rule)
Now, what do I do? Is this correct:
y'=cos(x+y)+cos(x+y)y'
y'-cos(x+y)y'=cos(x+y)
y'(1-cos(x+y))=cos(x+y)
y'= cos(x+y) / (1-cos(x+y) )
?
Yes, implicit differentiation is different from explicit differentiation. In implicit differentiation, we differentiate both the dependent and independent variables together, while in explicit differentiation, we differentiate only the dependent variable with respect to the independent variable.
Implicit differentiation is useful in scientific research as it allows us to find the derivative of a function that is not explicitly written in terms of the independent variable. This is particularly useful in situations where the relationship between variables is not easily expressed and can help in analyzing complex systems and phenomena.
Yes, implicit differentiation can be used to find second or higher order derivatives. We can continue to differentiate implicitly by following the same steps as we do for first-order derivatives, but with respect to the new derivative obtained in the previous step.
The key steps involved in using implicit differentiation include identifying the dependent and independent variables, differentiating both sides of the equation with respect to the independent variable, isolating the derivative, and solving for the desired derivative by substituting values for the independent and dependent variables.
Yes, there are some limitations to using implicit differentiation. It may not always be possible to isolate the derivative or solve for it explicitly. In such cases, we may need to resort to other methods or numerical approximations to find the derivative. Additionally, implicit differentiation may not work for all types of functions, such as those with discontinuities or non-differentiable points.