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

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Answer 1 · 2017-06-30T18:09:07

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

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

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

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

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

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

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