Question

Find the point on the curve y=x/(1+x^2) ,where the tangent to the curve has the greatest slope

Answer 1

Curve is given
y = x/(1 + x²)
Slope of tangent of curve y = x/(1 + x²) is dy/dx
So, differentiate y with respect to x
dy/dx = {dx/dx(1 + x²) - d(1+x²)/dx x}/(1 + x²)²
= {(1 + x²) - 2x²}/(1 + x²)
= (1 - x²)/(1 + x²)²

Hence, slope of tangent of curve y is (1 - x²)/(1 + x²)² = g(x) [let ]
We have to find the point, where tangent of slope is greatest .
So, differentiate dy/dx with respect to x
g'(x) = {(-2x)(1 + x²)² - 2x(1 + x²)(1 - x²)}/(1 + x²)²
= (1+ x²){-2x - 2x³ - 2x + 2x³ }/(1 + x²)²
= -4x/(1 + x²) , equate g'(x) with zero
then, -4x/(1 + x²) = 0⇒ x = 0 put x = 0 in y = x/(1 + x²) , y = 0
Now, check g"(x) < 0 at (0,0) if yes then (0,0) is the point on curve where slope of tangent of curve is greatest.
g"(x) = -4(1-x²)/(1+x²)² < 0 at (0,0)

Hence, answer is (0,0)

Answer 2

Answer:(0,0)

Step-by-step explanation:

Curve is given

y = x/(1 + x²)

Slope of tangent of curve y = x/(1 + x²) is dy/dx

So, differentiate y with respect to x

dy/dx = {dx/dx(1 + x²) - d(1+x²)/dx x}/(1 + x²)²

= {(1 + x²) - 2x²}/(1 + x²)

= (1 - x²)/(1 + x²)²

Hence, slope of tangent of curve y is (1 - x²)/(1 + x²)² = g(x) [let ]

We have to find the point, where tangent of slope is greatest .

So, differentiate dy/dx with respect to x

g'(x) = {(-2x)(1 + x²)² - 2x(1 + x²)(1 - x²)}/(1 + x²)²

= (1+ x²){-2x - 2x³ - 2x + 2x³ }/(1 + x²)²

= -4x/(1 + x²) , equate g'(x) with zero

then, -4x/(1 + x²) = 0⇒ x = 0 put x = 0 in y = x/(1 + x²) , y = 0

Now, check g"(x) < 0 at (0,0) if yes then (0,0) is the point on curve where slope of tangent of curve is greatest.

g"(x) = -4(1-x²)/(1+x²)² < 0 at (0,0)

Hence, answer is (0,0)