Partial derivaives
𝑑𝑦 𝑑𝑥 (4𝑥 2 cos(4y) + 3𝑒 3𝑦 ln(2𝑥) + 12𝑥 5 𝑦 2 )=−(20𝑥 4 𝑦 3 + 𝑒 3𝑦 𝑥
+ 2𝑥 sin(4𝑦))
𝑒 3𝑦 𝑥
20𝑥 4 𝑦 3 + + 2𝑥 sin(4𝑦) 4𝑥 2 cos(4y) + 3𝑒 3𝑦 ln(2𝑥) + 12𝑥 5 𝑦 2
𝑑𝑦 𝑑𝑥
=−
𝑑𝑦 𝑑𝑥
𝑑𝑦 𝑑𝑥
𝑓 𝑥 𝑓 𝑦 , but first we must rearrange for all variables to be
=−
Now, instead solve for
via the formula
on one side like so:
𝑥 2 sin(4𝑦) + 𝑒 3𝑦 ln(2𝑥) + 4𝑥 5 𝑦 3 −3=0
𝑒 3𝑦 𝑥
2𝑥 sin(4𝑦) + +20𝑥 4 𝑦 3 4𝑥 2 cos(4𝑦) + 3𝑒 3𝑦 ln(2𝑥) + 12𝑥 5 𝑦 2
𝑑𝑦 𝑑𝑥
=−
This illustrates that it is significantly more efficient to utilize partial derivatives in such problems rather than using implicit differentiation. Likewise, this formula holds for functions of more than one variable as is show below for functions of 3 variables 5 , where x and y are independent of each other.
𝑓(𝑥, 𝑦, 𝑧) = 0,
𝑧 = 𝑧(𝑥, 𝑦)
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑥
𝑑𝑓 𝑑𝑧
𝑓 𝑥
+𝑓 𝑦
+𝑓 𝑧
=0
𝑑𝑦 𝑑𝑥 =0 because x and y are independent variables and
𝑑𝑥 𝑑𝑥
=1 this gives:
Since
𝑑𝑓 𝑑𝑧
𝑓 𝑥 +𝑓 𝑧
=0
Rearranging this gives,
𝑑𝑧 𝑑𝑥
𝑓 𝑥 𝑓 𝑧
=−
The same principle can be applied when differentiating in respect to y to get the equation below.
𝑓 𝑦 𝑓 𝑧
𝑑𝑧 𝑑𝑦
=−
In general, an equation of any number of variables can be differentiated like this as long as 𝛼 1 , 𝛼 2 , 𝛼 3 ,…, 𝛼 𝑛−1 are independent variables and 𝛼 𝑛 = 𝛼 𝑛 (𝛼 1 , 𝛼 2 , 𝛼 3 ,…, 𝛼 𝑛−1 ) .
𝐺𝑖𝑣𝑒𝑛 𝑓(𝛼 1 , 𝛼 2 , 𝛼 3 ,…, 𝛼 𝑛−1 , 𝛼 𝑛 )=0 𝑑𝛼 𝑛 𝑑𝛼 𝑖 =− 𝑓 𝛼 𝑖 𝑓 𝛼 𝑛 , 1≤𝑖≤𝑛−1, 𝑖 ∈ℕ
Higher order partial derivatives
5 Dawkins (2003).
132
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