Video Transcript
Let 𝑧 be equal to 𝑓 of 𝑥, 𝑦
over the curve 𝑥 equals 𝑡 squared plus one, 𝑦 equals 𝑡 squared minus one. Write an expression for d𝑧 by d𝑡,
giving your answer in terms of the partial derivatives of 𝑓.
We see that 𝑧 is a multivariate
function. It’s a function in both 𝑥 and
𝑦. And we’re asked to find an
expression for d𝑧 by d𝑡 giving our answers in terms of the partial derivatives of
𝑓. Those are 𝜕𝑓 𝜕𝑥 — that’s the
partial derivative of 𝑓 with respect to 𝑥 — and 𝜕𝑓 𝜕𝑦 — the partial
derivative of 𝑓 with respect to 𝑦. Now, our functions for 𝑥 and 𝑦
are themselves differentiable functions of 𝑡. And so we can quote the chain rule
for one independent variable. Here, that’s 𝑡.
This says that suppose 𝑥 and 𝑦
are differentiable functions of 𝑡 and 𝑧 is a function in 𝑥 and 𝑦 as a
differentiable function. Then this means 𝑧 is 𝑓 of 𝑥 of
𝑡, 𝑦 of 𝑡 is a differentiable function of 𝑡. Such that d𝑧 by d𝑡, the
derivative of 𝑧 with respect to 𝑡 is equal to the partial derivative of 𝑧 with
respect to 𝑥 times the derivative of 𝑥 with respect to 𝑡 plus the partial
derivative of 𝑧 with respect to 𝑦 times the ordinary derivative of 𝑦 with respect
to 𝑡. Well, in this case, 𝑥 is the
function 𝑡 squared plus one and 𝑦 is the function 𝑡 squared minus one. These are both polynomials in 𝑡,
and we know polynomials are differentiable over their entire domain. So 𝑥 and 𝑦 are both
differentiable functions.
We’re going to evaluate d𝑥 by d𝑡,
the ordinary derivative of 𝑥 with respect to 𝑡, and d𝑦 by d𝑡, the ordinary
derivative of 𝑦 with respect to 𝑡. Now, we know that, to differentiate
a power term, we multiply the entire term by the exponent, then reduce that exponent
by one. So in both cases, the derivative of
𝑡 squared with respect to 𝑡 is two 𝑡. But we also know that the
derivative of a constant is zero. So both one and negative one
differentiate to zero. And we see that d𝑥 by d𝑡 and d𝑦
by d𝑡 are both equal to two 𝑡.
The question tells us to give our
answer in terms of the partial derivatives of 𝑓. Those are 𝜕𝑧 𝜕𝑥 and 𝜕𝑧
𝜕𝑦. And so we can say that the
derivative of 𝑧 with respect to 𝑡 is equal to 𝜕𝑧 𝜕𝑥 times two 𝑡 plus 𝜕𝑧
𝜕𝑦 times two 𝑡. We simplify a little, and we can
write this as two 𝑡 𝜕𝑧 𝜕𝑥 plus two 𝑡 𝜕𝑧 𝜕𝑦, where we said 𝜕𝑧 𝜕𝑥 is the
first partial derivative of 𝑧 with respect to 𝑥 and 𝜕𝑧 𝜕𝑦 is the first partial
derivative of 𝑧 with respect to 𝑦.
But remember, 𝑧 is a function in
𝑥 and 𝑦, where 𝑥 is 𝑡 squared plus one and 𝑦 is 𝑡 squared minus one. So we can rewrite 𝜕𝑧 𝜕𝑥 as
shown and 𝜕𝑧 𝜕𝑦 as shown. And so we found an expression for
d𝑧 by d𝑡 in terms of the partial derivatives of 𝑓. It’s two 𝑡 times the first partial
derivative of the function in 𝑡 squared plus one and 𝑡 squared minus one with
respect to 𝑥 plus two 𝑡 times the first partial derivative of the function in 𝑡
squared plus one and 𝑡 squared minus one with respect to 𝑦.
