# Video: Chain Rule for Multivariate Functions

The differentiable functions 𝑥 = ϕ(𝑡), 𝑦 = ψ(𝑡) describe a curve with (ϕ(1), ψ(1)) = (2, 3). Use the chain rule to find an expression for d𝑤/d𝑡|_(𝑡 = 1), where 𝑤 = 𝑒^(𝑥) cos (𝑦).

03:42

### Video Transcript

The differentiable functions 𝑥 equals ϕ of 𝑡, 𝑦 equals ψ of 𝑡 describe a curve with ϕ of one, ψ of one equals two, three. Use the chain rule to find an expression for d𝑤 by d𝑡 evaluated at 𝑡 equals one, where 𝑤 equals 𝑒 to the 𝑥 power times cos of 𝑦.

𝑤 is a multivariable function. It’s a function in terms of both 𝑥 and 𝑦. Now we’re told that the functions 𝑥 and 𝑦 are differentiable functions in 𝑡. And we’re told to use the chain rule to find an expression for the derivative of 𝑤 with respect to 𝑡. The chain rule says, suppose that 𝑥 is some function of 𝑡 and 𝑦 is also some function of 𝑡 and they’re differentiable functions and 𝑤 is some function of 𝑥 and 𝑦. Then 𝑤 equals 𝑓 of 𝑔 of 𝑡, ℎ of 𝑡 is a differentiable function of 𝑡 such that the derivative of 𝑤 with respect to 𝑡 is equal to the first partial derivative of 𝑤 with respect to 𝑥 times d𝑥 by d𝑡. Plus the first partial derivative of 𝑤 with respect to 𝑦 times d𝑦 by d𝑡.

Now both 𝜕𝑤, 𝜕𝑥 and 𝜕𝑤, 𝜕𝑦 are partial derivatives, whereas d𝑥 by d𝑡 and d𝑦 by d𝑡 are the ordinary derivatives of 𝑥 and 𝑦 with respect to 𝑡. We know that 𝑤 is equal to 𝑒 to the power of 𝑥 times cos of 𝑦. So let’s find the partial derivative of 𝑤 with respect to 𝑥 and the partial derivative of 𝑤 with respect to 𝑦.

Now remember, when a function is made up of more than one variable, we can see how the function changes as we let just one of those variables change and hold all the others constant. For 𝜕𝑤, 𝜕𝑥, we’re going to let 𝑥 change and we’re therefore going to treat 𝑦 as a constant. Now if 𝑦 is a constant, cos of 𝑦 is also a constant. And when we differentiate a constant times 𝑒 to the power of 𝑥, we just get that constant times 𝑒 to the power of 𝑥. So the first partial derivative of 𝑤 with respect to 𝑥 — 𝜕𝑤, 𝜕𝑥 — is 𝑒 to the power of 𝑥 cos 𝑦.

However, when finding 𝜕𝑤, 𝜕𝑦 — that’s the first partial derivative of 𝑤 with respect to 𝑦 — we vary 𝑦 and we keep 𝑥 constant. Now if we’re imagining 𝑥 is a constant, then 𝑒 to the power of 𝑥 is also a constant. And so we know that when we differentiate a constant times cos of 𝑦 with respect to 𝑦, we get negative that constant times sin of 𝑦. So 𝜕𝑤, 𝜕𝑦 is negative 𝑒 to the power of 𝑥 times sin of 𝑦. This means d𝑤 by d𝑡 is equal to 𝜕𝑤, 𝜕𝑥, which is 𝑒 to the power of 𝑥 cos of 𝑦 times the derivative of 𝑥 with respect to 𝑡.

Now remember, 𝑥 is ϕ of 𝑡. So the derivative of 𝑥 with respect to 𝑡 is ϕ prime of 𝑡. Then we add 𝜕𝑤, 𝜕𝑦, which is negative 𝑒 to the power of 𝑥 sin 𝑦, times d𝑦 by d𝑥. And remember, 𝑦 is ψ of 𝑡. So that’s ψ prime of 𝑡.

The question’s asking us to find an expression for d𝑤 by d𝑡 at 𝑡 equals one. And we know when 𝑡 is equal to one, 𝑥 is equal to two and 𝑦 is equal to three. So 𝑒 to the power of 𝑥 cos 𝑦 becomes 𝑒 squared times cos of three, whereas ϕ prime of 𝑡 becomes ϕ prime of one. Then our second term is negative 𝑒 squared sin of three times ψ prime of one.

And of course, since we don’t know the functions ϕ and ψ, we’re done. d𝑤 by d𝑡 evaluated at 𝑡 equals one is equal to 𝑒 squared cos of three ϕ prime of one minus 𝑒 squared sin of three ψ prime of one.