Partial derivaives
𝑑𝑓 = 𝑓(𝑥 + 𝑑𝑥, 𝑦 + 𝑑𝑦) − 𝑓(𝑥, 𝑦)
𝑑𝑓 = [𝑓(𝑥 + 𝑑𝑥, 𝑦 + 𝑑𝑦) − 𝑓(𝑥, 𝑦 + 𝑑𝑦)] + [𝑓(𝑥, 𝑦 + 𝑑𝑦) − 𝑓(𝑥, 𝑦)]
From our previous definition of partial derivatives, we can rewrite this equation in terms of partial derivatives. Taking the limits 𝑑𝑥 →0 and 𝑑𝑦 →0 , results in a first order approximation 3 where only the first term of each sequence is left.
𝜕 2 𝑓 𝜕𝑥 2
𝜕 3 𝑓 𝜕𝑥 3
𝜕 2 𝑓 𝜕𝑦 2
𝜕 3 𝑓 𝜕𝑦 3
𝜕𝑓 𝜕𝑥
𝜕𝑓 𝜕𝑦
𝑑𝑥 2 +
𝑑𝑥 3 …)+(
𝑑𝑦 2 +
𝑑𝑦 3 …))
lim 𝑑𝑥→0 𝑑𝑦→0
𝑑𝑓 = lim 𝑑𝑥→0 𝑑𝑦→0
((
𝑑𝑥 +
𝑑𝑦+
𝜕𝑓 𝜕𝑥
𝜕𝑓 𝜕𝑦
𝑑𝑓 =
𝑑𝑥+
𝑑𝑦
Dividing the equation by the differential we want to differentiate the function in respect to gives:
𝜕𝑓 𝜕𝑡
𝜕𝑓 𝜕𝑥
𝜕𝑥 𝜕𝑡
𝜕𝑓 𝜕𝑦
𝜕𝑦 𝜕𝑡
=
+
The same can be done with s , x , y or any other variable – albeit differentiating with respect to variables not defined in the function will result as a 0. Using this formula, we can now find 𝜕𝑓 𝜕𝑡 like so:
𝜕𝑓 𝜕𝑡
tan(2𝑥) 𝑦
𝑡 √2𝑠 +𝑡 2
=(2𝑠𝑒𝑐 2 (2𝑥) ln(3𝑦))(3𝑡 2 )+(
)(
)
We can then rewrite this in only s and t which gives:
tan(2𝑡 3 −2𝑠 2 )×𝑡 2𝑠 +𝑡 2
𝜕𝑓 𝜕𝑡 =6𝑡 2 𝑠𝑒𝑐 2 (2𝑡 3 −2𝑠 2 ) ln (3√2𝑠 + 𝑡 2 )+
One a separate note, partial derivatives can also be used instead of implicit differentiation to solve problems. Below is derived a formula for 𝑑𝑦 𝑑𝑥 in terms of partial derivatives 4 for two variable functions.
𝑓(𝑥, 𝑦) = 0,
𝑦 = 𝑦(𝑥)
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑥
𝑓 𝑥
+𝑓 𝑦
=0
𝑑𝑥 𝑑𝑥
=1 , this gives:
Since
𝑑𝑦 𝑑𝑥
𝑓 𝑥 +𝑓 𝑦
=0
𝑑𝑦 𝑑𝑥 the subject of the equation gives:
Making
𝑑𝑦 𝑑𝑥
𝑓 𝑥 𝑓 𝑦
=−
Take a look at the example below where implicit differentiation is used to solve for 𝑑𝑦 𝑑𝑥 : 𝑥 2 sin(4𝑦) + 𝑒 3𝑦 ln(2𝑥) = 3 − 4𝑥 5 𝑦 3
𝑒 3𝑦 𝑥
dy dx
𝑑𝑦 𝑑𝑥
𝑑𝑦 𝑑𝑥
2𝑥 sin(4𝑦) + 4𝑥 2 cos(4y)
+3𝑒 3𝑦 ln(2𝑥)
=−20𝑥 4 𝑦 3 −12𝑥 5 𝑦 2
+
3 Wikipedia n.d. 4 Dawkins (2003).
131
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