The Weak Chain

Calculus Level pending

Let z = x y z = xy , where x = f ( u ) x= f(u) and y = g ( u ) y = g(u) . Suppose that f f and g g are differentiable, such that f ( 1 ) = 2 , g ( 1 ) = − 2 , d x d u ( 1 ) = − 1 f(1) = 2, \; g(1) = -2, \; \dfrac{dx}{du}(1) = -1 and d y d u ( 1 ) = 5 \dfrac{dy}{du}(1) = 5 .

Evaluate d z d u ( 1 ) \dfrac{dz}{du}(1) .


The answer is 12.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Applying the chain rule, we have that

d z d u = ∂ z ∂ x d x d u + ∂ z ∂ y d y d u ⇔ d z d u = g ( u ) f ′ ( u ) + f ( u ) g ′ ( u ) \dfrac{dz}{du} = \dfrac{\partial z}{\partial x} \dfrac{dx}{du} + \dfrac{\partial z}{\partial y} \dfrac{dy}{du} \Leftrightarrow \dfrac{dz}{du} = g(u) f'(u) + f(u) g'(u)

Let u = 1 u = 1 . Then d z d u ( 1 ) = − 2 ⋅ − 1 + 2 ⋅ 5 = 12. \dfrac{dz}{du}(1) = -2 \cdot -1 + 2 \cdot 5 = \boxed{12.} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

z = x y d z d u = d x d u ⋅ y + x ⋅ d y d u d z d u ∣ u = 1 = d x d u ⋅ y ∣ u = 1 + x ⋅ d y d u ∣ u = 1 = ( − 1 ) ( − 2 ) + 2 ( 5 ) = 2 + 10 = 12 \begin{aligned} z & = xy \\ \frac {dz}{du} & = \frac {dx}{du} \cdot y + x \cdot \frac {dy}{du} \\ \frac {dz}{du}\bigg|_{u=1} & = \frac {dx}{du} \cdot y\bigg|_{u=1} + x \cdot \frac {dy}{du} \bigg|_{u=1} \\ & = (-1)(-2) + 2(5) \\ & = 2+10 \\ & = \boxed{12} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...