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.?

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)#

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)#.