Question

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.?

Answer 1

`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)`

Answer 2

`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)`.