A curve has the equation y=(x+2)sqrt(x-1). show that dy/dx=(kx)/(sqrt(x-1)), where k is a contant and state the value of k.?

2 Answers
Mar 17, 2018

k=3/2

Explanation:

By using the rule for differentiating a product, we get :

dy/dx = (d/dx (x+2)) sqrt(x-1) + (x+2) d/dx (sqrt(x-1))
qquad = 1times sqrt(x-1) + (x+2) times 1/(2sqrt(x-1))
qquad = (2*(sqrt(x-1))^2+(x+2))/(2sqrt(x-1)) =(2x-2+x+2)/(2sqrt(x-1))
qquad = (3/2x)/sqrt(x-1)

Mar 17, 2018

dy/dx = (3/2 x)/sqrt(x-1) => k =3/2.

Explanation:

We have y = (x+2)sqrt(x-1).

By the product rule, which states that if f(x) = g(x)h(x) then

color(blue)(f'(x) = g(x)h'(x) + g'(x)h(x).

y' = (x+2)[sqrt(x-1)]' + [x+2]'sqrt(x-1)

We have to find color(red)([sqrt(x-1)]' and color(red)([x+2]'.

Since 2 is a constant, [x+2]' = [x]' = 1.

For [sqrt(x-1)]', you have to apply the power rule.

color(blue)([f(x)^n]' = nf(x)^(n-1) * f'(x).

In this case, n = 1/2 and f(x) = x-1.

=> color(red)([sqrt(x-1)]' = 1/(2sqrt(x-1))

Finally, the derivative of y is

dy/dx = y' = (x+2)/(2sqrt(x-1))+ sqrt(x-1)

Multiply both the numerator and denominator on the fraction

sqrt(x-1)/1 by 2sqrt(x-1).

dy/dx = (x+2)/(2sqrt(x-1)) + (2x-2)/(2sqrt(x-1))

color(red)(dy/dx = (3x)/(2sqrt(x-1)) = (color(blue)(3/2) x)/sqrt(x-1)

=> color(blue)(k=3/2).