How do I differentiate $\cos(x+y)$?

In summary, to find $\frac{dy}{dx}$ when $y^2+\cos(x+y) = 1$, use the chain rule and substitute $u(x)=x+y$ to get $-\sin(x+y)(1+\frac{dy}{dx})=2y\d{y}{x}$, then solve for $\frac{dy}{dx}$.
  • #1
Guest2
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If $y^2+\cos(x+y) = 1$ find $\frac{dy}{dx}$. How do I differentiate $\cos(x+y)$ bit?
 
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  • #2
Guest said:
If $y^2+\cos(x+y) = 1$ find $\frac{dy}{dx}$. How do I differentiate $\cos(x+y)$ bit?

Use the chain rule:

\(\displaystyle \frac{d}{dx}\left(\cos(u(x))\right)=-\sin(u(x))\d{u}{x}\)

If $u(x)=x+y$, then what do you get? :)
 
  • #3
MarkFL said:
Use the chain rule:

\(\displaystyle \frac{d}{dx}\left(\cos(u(x))\right)=-\sin(u(x))\d{u}{x}\)

If $u(x)=x+y$, then what do you get? :)
I get $-\sin(x+y)(1+\frac{dy}{dx})$
 
  • #4
Guest said:
I get $-\sin(x+y)(1+\frac{dy}{dx})$

Yes, that's correct. (Yes)

So, then you have:

\(\displaystyle 2y\d{y}{x}-\sin(x+y)\left(1+\d{y}{x}\right)=0\)

And you just need to solve for \(\displaystyle \d{y}{x}\). :)
 
  • #5
MarkFL said:
Yes, that's correct. (Yes)

So, then you have:

\(\displaystyle 2y\d{y}{x}-\sin(x+y)\left(1+\d{y}{x}\right)=0\)

And you just need to solve for \(\displaystyle \d{y}{x}\). :)
Done! Many thanks! :D
 

FAQ: How do I differentiate $\cos(x+y)$?

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly written in terms of one variable. It allows us to find the rate of change of a dependent variable with respect to an independent variable, even when the dependent variable is not expressed as a function of the independent variable.

How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used to find the derivative of a function that is written explicitly in terms of one variable, while implicit differentiation is used to find the derivative of a function that is not explicitly written in terms of one variable. In implicit differentiation, the dependent variable is usually written in terms of both the independent variable and the dependent variable, whereas in explicit differentiation, the dependent variable is only written in terms of the independent variable.

What are the main steps involved in implicit differentiation?

The main steps involved in implicit differentiation are: 1) Differentiating both sides of the equation with respect to the independent variable, 2) Using the chain rule to find the derivative of any terms that involve the dependent variable, 3) Isolating the derivative of the dependent variable on one side of the equation, and 4) Simplifying the resulting expression, if necessary.

Why is implicit differentiation useful?

Implicit differentiation allows us to find the derivative of a function without having to explicitly solve for one variable in terms of the other. This is especially useful when dealing with complex functions that cannot be easily solved for one variable. It is also used in applications such as optimization and curve sketching.

What are some common mistakes to avoid in implicit differentiation?

Some common mistakes to avoid in implicit differentiation include: 1) Forgetting to apply the chain rule when differentiating terms that involve the dependent variable, 2) Forgetting to use the product rule or quotient rule when differentiating products or quotients, 3) Not properly isolating the derivative of the dependent variable, and 4) Making errors in algebraic simplification. It is important to double check each step and be mindful of the rules of differentiation.

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