Explicit Differentiation
Explicit function: If [katex]y[/katex] is easily expressed in term of the independent variables [katex]x[/katex],Then [katex]y[/katex] is called an Explicit function of [katex]x[/katex]. Symbolically it is written as [katex]y=f(x)[/katex]
Examples:
[katex]y+3x-5=0[/katex]
[katex]y^5+2x^3-3=0[/katex]
[katex]y^2=\left(x+4\right)^\frac12[/katex]
Procedure:
- Step (1) when [katex]x[/katex] and [katex]y[/katex] are not amalgamated or Explicit we assumethat [katex]y[/katex] is differentiable function of [katex]x[/katex].
- Step (2) Differentiate both sides of eq w.r.t [katex]x[/katex].
- Step (3) Solve the resulting eq for [katex]\frac{dy}{dx}[/katex].
Example 1:
Explicit Differentiation
[katex]y+3x-5=0[/katex]
Differentiate w.r.t [katex]x[/katex]
[katex]\frac d{dx}\left(y+3x-5\right)=\frac d{dx}0[/katex]
[katex]\frac d{dx}y+\frac d{dx}3x-\frac d{dx}5=0[/katex]
[katex]\frac d{dx}y+3\frac d{dx}x-0=0[/katex]
[katex]\frac d{dx}y+3=0[/katex]
[katex]\boxed{\frac d{dx}y=-3}[/katex]
Example 2:
Explicit Differentiation.
[katex]y^5+2x^3-3=0[/katex]
Differentiate w.r.t [katex]x[/katex]
[katex]\frac d{dx}\left(y^5+2x^3-3\right)=\frac d{dx}0[/katex]
[katex]\frac d{dx}y^5+\frac d{dx}2x^3-\frac d{dx}3=0[/katex]
[katex]5y^4\frac{dy}{dx}+2\frac d{dx}x^3-0=0[/katex]
[katex]5y^4\frac{\displaystyle dy}{\displaystyle dx}+2(3x^2)=0[/katex]
[katex]5y^4\frac{\displaystyle dy}{\displaystyle dx}+6x^2=0[/katex]
[katex]5y^4\frac{\displaystyle dy}{\displaystyle dx}=-6x^2[/katex]
[katex]\boxed{\frac{\displaystyle dy}{\displaystyle dx}=\frac{-6x^2}{5y^4}}[/katex]
Example 3:
Explicit Differentiation
[katex]y^2=\left(x+4\right)^\frac12[/katex]
Differentiate w.r.t [katex]x[/katex]
[katex]\frac d{dx}y^2=\frac d{dx}\left(x+4\right)^\frac12[/katex]
[katex]2y\frac d{dx}y=\frac12\left(x+4\right).^{\frac12-1}\frac d{dx}\left(x+4\right)[/katex]
[katex]2y\frac d{dx}y=\frac12\left(x+4\right).^{-\frac12}\left[\frac d{dx}x+\frac d{dx}4\right][/katex]
[katex]2y\frac d{dx}y=\frac12\left(x+4\right).^{-\frac12}\left[1+0\right][/katex]
[katex]2y\frac d{dx}y=\frac12\left(x+4\right).^{-\frac12}[/katex]
[katex]\frac d{dx}y=\frac{\frac12\left(x+4\right).^{-\frac12}}{2y}[/katex]
[katex]\frac d{dx}y=\frac{\left(x+4\right).^{-\frac12}}{4y}[/katex]
[katex]\boxed{\frac d{dx}y=\frac1{4y.\left(x+4\right).^{-\frac12}}}[/katex]