How do you find the derivative of #y=[(2-3x^2)^(1/2)](5x+2)#?

1 Answer
Jun 30, 2017

#(-15x^2-15x+4)/sqrt(2-3x^2)#

Explanation:

The derivative of a product #color(red)f(x)*color(green)g(x)# is:

#f'(x)*g(x)+f(x)*g'(x)#

Let's assume that

#color(red)(f(x)=(2-3x^2)^(1/2))#

and

#color(green)(g(x)=5x+2)#

Since

#f(x)=(f_1(x))^a->f'(x)=color(red)(a(f_1(x))^(a-1))*color(blue)(f_1'(x))#

the derivative finally is:

#color(red)(1/cancel2(2-3x^2)^(1/2-1)*color(blue)((-3*cancel2x)))*color(green)((5x+2))+color(red)((2-3x^2)^(1/2))*color(green)5#

#=-3*(2-3x^2)^(-1/2)* (5x+2)+5*(2-3x^2)^(1/2)#

or

#-(3(5x+2))/sqrt(2-3x^2)+5sqrt(2-3x^2)#

#=(-3(5x+2)+5(2-3x^2))/sqrt(2-3x^2)#

#=(-15x-6+10-15x^2)/sqrt(2-3x^2)#

#=(-15x^2-15x+4)/sqrt(2-3x^2)#