What is the implicit derivative of #4= (x+y)^2 #?

Question

What is the implicit derivative of #4=   (x+y)^2 #?

Paisley Johnson · Accepted Answer

You can use calculus and spend a few minutes on this problem or you can use algebra and spend a few seconds, but either way you'll get #dy/dx=-1#.

Begin by taking the derivative with respect to both sides: #d/dx(4)=d/dx(x+y)^2# On the left, we have the derivative of a constant - which is just #0#. That breaks the problem down to: #0=d/dx(x+y)^2# To evaluate #d/dx(x+y)^2#, we need to use the power rule and the chain rule: #d/dx(x+y)^2=(x+y)'*2(x+y)^(2-1)# Note: we multiply by #(x+y)'# because the chain rule tells us we have to multiply the derivative of the whole function (in this case #(x+y)^2# by the inside function (in this case #(x+y)#). #d/dx(x+y)^2=(x+y)'*2(x+y)# As for #(x+y)'#, notice that we can use the sum rule to break it into #x'+y'#. #x'# is simply #1#, and because we don't actually know what #y# is, we have to leave #y'# as #dy/dx#: #d/dx(x+y)^2=(1+dy/dx)(2(x+y))# Now that we've found our derivative, the problem is: #0=(1+dy/dx)(2(x+y))# Doing some algebra to isolate #dy/dx#, we see: #0=(1+dy/dx)(2x+2y)# #0=2x+dy/dx2x+dy/dx2y+2y# #0=x+dy/dxx+dy/dxy+y# #-x-y=dy/dxx+dy/dxy# #-x-y=dy/dx(x+y)# #dy/dx=(-x-y)/(x+y)# Interestingly, this equals #-1# for all #x# and #y# (except when #x=-y#). Therefore, #dy/dx=-1#. We could have actually figured this out without using any calculus at all! Look at the equation #4=(x+y)^2#. Take the square root of both sides to get #+-2=x+y#. Now subtract #x# from both sides, and we have #y=+-2-x#. Remember these from algebra? The slope of this line is #-1#, and since the derivative is the slope, we could have just said #dy/dx=-1# and avoided all that work.

Evelyn Agard · Answer

undefined To find the implicit derivative of $4 = (x + y)^2$, you differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. The derivative of $4$ with respect to $x$ is $0$.

Differentiating $(x + y)^2$ with respect to $x$ using the chain rule and the power rule, we get:

$$
\frac{d}{dx}[(x + y)^2] = 2(x + y) \frac{d}{dx}(x + y)
$$

Applying the chain rule to $\frac{d}{dx}(x + y)$, we have:

$$
\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}
$$

Substituting this back into our expression, we get:

$$
2(x + y)(1 + \frac{dy}{dx}) = 0
$$

Now, solve for $\frac{dy}{dx}$:

$$
\begin{align*}
2(x + y) \left(1 + \frac{dy}{dx}\right) &= 0 \
2(x + y) + 2(x + y) \frac{dy}{dx} &= 0 \
2(x + y) \frac{dy}{dx} &= -2(x + y) \
\frac{dy}{dx} &= \frac{-2(x + y)}{2(x + y)} \
\frac{dy}{dx} &= -1
\end{align*}
$$

Therefore, the implicit derivative of $4 = (x + y)^2$ with respect to $x$ is $-1$.