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

Question

Aurora Alvarado · Accepted Answer

#(-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)#

Assume for now 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))#

At last, the derivative 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)#

William Abdalla · Answer

undefined To find the derivative of $ y = [(2 - 3x^2)^{\frac{1}{2}}](5x + 2) $, you can use the product rule and the chain rule. First, differentiate each part separately, then apply the product rule. The derivative of $ (2 - 3x^2)^{\frac{1}{2}} $ with respect to $ x $ involves the chain rule. The derivative of $ 5x + 2 $ is straightforward:

$$ \frac{d}{dx}(5x + 2) = 5 $$

Now, let's find the derivative of $ (2 - 3x^2)^{\frac{1}{2}} $:

$$ \frac{d}{dx}[(2 - 3x^2)^{\frac{1}{2}}] = \frac{1}{2}(2 - 3x^2)^{-\frac{1}{2}}(-6x) $$

Combining the derivatives using the product rule:

$$ y' = \left[(2 - 3x^2)^{\frac{1}{2}}\right](5) + \left[(5x + 2)\right]\left[\frac{1}{2}(2 - 3x^2)^{-\frac{1}{2}}(-6x)\right] $$

$$ y' = 5\sqrt{2 - 3x^2} - 3x(5x + 2)\frac{1}{\sqrt{2 - 3x^2}} $$

$$ y' = 5\sqrt{2 - 3x^2} - \frac{15x^2 + 6x}{\sqrt{2 - 3x^2}} $$

Thus, the derivative of $ y = [(2 - 3x^2)^{\frac{1}{2}}](5x + 2) $ is $ y' = 5\sqrt{2 - 3x^2} - \frac{15x^2 + 6x}{\sqrt{2 - 3x^2}} $.