Differentiation of Implicit Function

Here you will learn what is the differentiation of implicit function with examples.

Let’s begin –

Differentiation of Implicit Function

If the variables x and y are connected by a relation of the form f(x,y) = 0 and it is not possible or convenient to express y as a function x in the form y  = \(\phi (x)\), then y is said to be an implicit function of x. To find \(dy\over dx\) in such a case, we differentiate both sides of the given relation with respect to x, keeping in mind that the derivative of \(\phi (y)\) with respect to x is \(d\phi\over dy\).\(dy\over dx\).

for example, \(d\over dx\) (sin y) = cos y. \(dy\over dx\) , \(d\over dx\) (\(y^2\)) = 2y\(dy\over dx\)

It should be noted that \(d\over dy\) (sin y) = cos y but \(d\over dx\) (sin y) = cos y. \(dy\over dx\).

Similarly , \(d\over dy\) \((y^3)\) = \(3y^2\) whereas \(d\over dx\) \((y^3)\) = \(3y^2\) \(dy\over dx\).

Example : If \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\) = 0, find \(dy\over dx\).

Solution : We have,

\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\) = 0

Differentiating both sides of this with respect to x, we get

\(d\over dx\) \((ax^2)\) + \(d\over dx\)(2hxy) + \(d\over dx\)\((by^2)\) + \(d\over dx\)(2gx) + \(d\over dx\)(2fy) + \(d\over dx\)(c) = \(d\over dx\)(0)

\(\implies\) a\(d\over dx\)\((x^2)\) + 2h\(d\over dx\)(xy) + b\(d\over dx\)\((y^2)\) + 2g\(d\over dx\)(x) + 2f\(d\over dx\)(y) + 0 = 0

\(\implies\) 2ax + 2h\((x{dy\over dx} + y)\) + b 2y\(dy\over dx\) + 2g.1 + 2f.\(dy\over dx\) = 0

= \(dy\over dx\)(2hx + 2by + 2f) + 2ax + 2hy + 2g = 0

\(dy\over dx\) = -\(2(ax + hy + g)\over 2(hx + by + f)\)

\(dy\over dx\) = -\((ax + hy + g)\over (hx + by + f)\)

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