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what is implicit differentiation and how is it achieved?

First off, an explicit definition needs to be understood. An explicit definition is the standard form in which many equations are found:

y = f(x)        (1)

for example:  y=x2+2

Here, all values of y can be explicitly defined by a function of x. An implicit expression is one where the equation cannot easily be rearranged/not be rearranged into a form similar to the one found in (1) where y can be expressed as a function of x.

In these situations the expression is said to be implicitly defined.

Implicit differentiation is the method in which these equations can be correctly and easily differentiated.

 

in implicit differentiation, y terms are considered to be a function of a function, or more simply:

 

y => f(y)

 

thus when differentiating a y term, the chain rule (or the rule for differentiating a function of a function) needs to be used.

dy/dx=dy/du * du/dx                (2)

                                                                                              

also when differentiating y in terms of x, or 

d/dx(f(y))=d/x(y) = dy/dx         (3)

 

so for example lets look at the following equation:

16y3 + 9x2y - 54x = 0           (4)

                                                

lets focus on the first term in this equation to start with 16y3

let A = 16y3

d/dx(A)=d/dx(16y3)                  (5)

here we have to use the chain rule. 

here u=y, thus looking back at (2) our terms are:

dy/du = d/du(16u3) = 48u2         (6)

du/dx = d/dx(y)  = dy/dx            (7)

thus combining (6) and (7) to get the answer for (5) you get:

d/dx(A) = 48u* dy/dx = dy/dx(48y2)            (8)

as you can see from (8), the desired term dy/dx is generated but the y term is still present, this is because y cannot be explicitly defined and this the gradient of the line is dependent on both x and y terms.

following this through for the next two terms in (4) you get:

 

for 9x2y, product rule is required. note that as this term includes an x term, these terms are differentiated normally.

d/dx(B) = d/dx(9x2y) = d/dx(9x2)*y + d/dx(y​)*9x= 18xy + dy/dx(9x2)

 

for -54x, this is simple differentiation:

d/dx(C) = d/dx(-54x) = -54

 

collecting the 3 differentiated terms of A,B and C, you get the answer to:

d/dx(16y3 + 9x2y - 54x) =  d/dx(A) + d/dx(B) + d/dx(C) = dy/dx(48y2) + 18xy + dy/dx(9x2) - 54 = 0        (9)

 

from here (9) can have its terms collected and rearranged to give the equation in terms of dy/dx as desired.

dy/dx(48y2 + 9x) = 54-18xy

dy/dx=(54-18xy)/(48y2 + 9x)           (10)

 

(note this can be further simplified by noticing that all constants in the expression are divisible by 3)

final answer would be:

dy/dx=(18-6xy)/(16y2 + 3x)              (11)

Jake S. GCSE Maths tutor, A Level Maths tutor, GCSE Electronics tutor...

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