The gradient of the normal to the curve at point (x, y) is, (√(1-x)) (1 + x)^(3/2)
What is gradient of a line?
"The gradient of a line is the measure of the steepness of a straight line. It is also known as slope of the line/curve."
What is normal?
"The normal is a line at right angles to the tangent."
What is tangent?
"It is a line that touches a curve at a single point and does not cross through it."
Quotient rule of differentiation:
[tex]\frac{d}{dx}[ {\frac{f(x)}{g(x)} }]=\frac{g(x)\frac{d}{dx} [f(x)]-f(x)\frac{d}{dx} [g(x)]}{(g(x))^2}[/tex]
For given example,
Given : (√(1-x))/(√(1+x))
[tex]y=\frac{\sqrt{1-x}}{\sqrt{1+x} }[/tex]
First we differentiate an expression with respect to 'x'.
[tex]\frac{dy}{dx}=\frac{d}{dx}(\frac{\sqrt{1-x}}{\sqrt{1+x} } )[/tex]
Using the Quotient Rule,
[tex]\Rightarrow \frac{dy}{dx} =\frac{\sqrt{1+x}\frac{d}{dx} [\sqrt{1-x} ]-\sqrt{1-x} \frac{d}{dx} [\sqrt{1+x} ]}{(\sqrt{1+x})^2 } \\\\\Rightarrow \frac{dy}{dx} =\frac{(\sqrt{1+x}(\frac{1}{2} (1-x)^{\frac{1}{2}-1})\frac{d}{dx} [1-x])-(\sqrt{1-x} (\frac{1}{2} (1+x)^{\frac{1}{2}-1})\frac{d}{dx} [\sqrt{1+x} ])}{(\sqrt{1+x})^2 } \\\\\Rightarrow \frac{dy}{dx} =-\frac{1}{\sqrt{1-x} (x+1)^{\frac{3}{2} }}[/tex]
This means the the gradient of the tangent to the curve is [tex]-\frac{1}{\sqrt{1-x} (x+1)^{\frac{3}{2} }}[/tex]
We know,
For a curve
gradient of the tangent at (x, y) × gradient of the normal at (x, y) = -1
⇒ the gradient of the normal to the curve at point (x, y) = [tex]\frac{-1}{(-\frac{1}{\sqrt{1-x} (x+1)^{\frac{3}{2} }})}[/tex]
⇒ the gradient of the normal to the curve at point (x, y) =
(√(1-x)) (1 + x)^(3/2)
Therefore, the gradient of the normal to the curve at point (x, y) is (√(1-x)) (1 + x)^(3/2)
Learn more about tangent and normal here:
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