## Continuity and Differentiability

# Derivatives of Functions in Parametric Forms

Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed via a third variable. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the form x = *f*(*t*), y = *g*(*t*) is said to be parametric form with* t* as a parameter.

In order to find derivative of function in such form, we have by chain rule.

\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}

\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\tt \left(whenever\ \frac{dx}{dt}\neq 0\right)

Thus \frac{dy}{dx}=\frac{g'(t)}{f'(t)}\ \left(as\ \frac{dy}{dt}=g'(t)\ and\ \frac{dx}{dt}=f'(t)\right) [provided f'(t) ≠ 0]

### Part1: View the Topic in this video From 49:18 To 53:58

### Part2: View the Topic in this video From 00:40 To 05:53

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- If y=f(x)+\frac{1}{y} then \frac{dy}{dx}=\frac{y^{2}f'(x)}{y^{2}+1}
- If f(x + y) = f(x) f(y) ∀ x, y ∈ R and f(x) ≠ 0 f(a) = k, f'(0) exists, then f'(a) = kf'(0).
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- \frac{d}{dx}\left[\tan^{-1}\left(\frac{a\cos f(x)-b \sin f(x)}{b\cos f(x)+a \sin f(x)}\right)\right]=f'(x)
- If \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y) then \frac{dy}{dx}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}
- If \sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a(x^3-y^3) then \frac{dy}{dx}=\frac{x^{2}}{y^{2}}\sqrt{\frac{1-y^{6}}{1-x^{6}}}
- If x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=1, then \frac{dy}{dx}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}
- If x
^{m}. y^{n}= a^{m + n}then \frac{dy}{dx}=\frac{-my}{nx} - If f(x, y) = 0 is a homogeneous equation in x and y then \frac{dy}{dx}=\frac{y}{x}.
- If f(x + y) = f(x) f(y) ∀ x, y ∈ R then f'(x) = f(x) + f'(0)
- If ax
^{2}+ 2hxy + by^{2}+ 2gx + 2fy + c = 0 then \frac{dy}{dx}=\frac{-(ax+hy+g)}{(hx+by+f)}