Find the equation of the tangent line to the curve

Question

y=x sqrt{x} at the point (16,64)?

 

 

Answer 1

The curve equation is y = x √x 

y = x^3/2

Differentiating on each side with respect of x .

y' = 3/2 [x^{(3/2) - 1} ] 

y' = 3/2 [x^1/2]

y' = (3√x)/2

Substitute the values (x , y ) = (16, 64) in above equation.

y' = (3√16)/2

y' = 6

This is the slope of tangent line to the curve at (16, 64).

To find the tangent line equation, substitute the values of m = 6 and (x , y ) = (16, 64)  in the slope intercept form of an equation y = mx + b.

64 = 6(16) + b

64 = 96 + b

b = - 96 + 64

b = - 32

Substitute m = 6 and b = -32 in y = mx + b.

Tangent line is y = 6x - 32.