How do you differentiate #y=(x^2+1)root3(x^2+2)#?

Question

Samantha Adkison · Accepted Answer

#y'=(2x(3(x^2+2)^(1/3)+(x^2+1)(x^2+2)^(-2/3)))/3# We have #y=(x^2+1)(x^2+2)^(1/3)# #y'=d/(dx)[(x^2+1)(x^2+2)^(1/3)]# #color(white)(X)=d/(dx)[x^2+1] (x^2+2)^(1/3)+d/(dx)[(x^2+2)^(1/3)] (x^2+1)# #color(white)(X)=(d/(dx)[x^2]+d/(dx)[1] )(x^2+2)^(1/3)+(d/(dx)[x^2+2] (x^2+2)^(1/3-1)*1/3) (x^2+1)# #color(white)(X)=(2x+0)(x^2+2)^(1/3)+(((d/(dx)[x^2]+d/(dx)[2]) (x^2+2)^(-2/3))/3) (x^2+1)# #color(white)(X)=2x(x^2+2)^(1/3)+(((2x+0) (x^2+2)^(-2/3))/3) (x^2+1)# #color(white)(X)=2x(x^2+2)^(1/3)+((2x(x^2+2)^(-2/3))/3) (x^2+1)# #color(white)(X)=2x(x^2+2)^(1/3)+(2x(x^2+1)(x^2+2)^(-2/3))/3# #color(white)(X)=(6x(x^2+2)^(1/3))/3+(2x(x^2+1)(x^2+2)^(-2/3))/3# #color(white)(X)=(6x(x^2+2)^(1/3)+2x(x^2+1)(x^2+2)^(-2/3))/3# #color(white)(X)=(2x(3(x^2+2)^(1/3)+(x^2+1)(x^2+2)^(-2/3)))/3#

William Abdalla · Answer

undefined To differentiate $ y = (x^2 + 1)\sqrt{3}(x^2 + 2) $, you can use the product rule. The product rule states that if $ y = u \cdot v $, where $ u $ and $ v $ are functions of $ x $, then $ \frac{dy}{dx} = u'v + uv' $.

Let $ u = x^2 + 1 $ and $ v = \sqrt{3}(x^2 + 2) $.

Now, find the derivatives of $ u $ and $ v $:

$$ u' = 2x $$
$$ v' = \frac{d}{dx}[\sqrt{3}(x^2 + 2)] = \sqrt{3}(2x) $$

Apply the product rule:

$$ \frac{dy}{dx} = (x^2 + 1) \cdot \sqrt{3}(2x) + 2x \cdot \sqrt{3}(x^2 + 2) $$

$$ = 2x\sqrt{3}(x^2 + 1) + 2x\sqrt{3}(x^2 + 2) $$

$$ = 2x\sqrt{3}(x^2 + 1 + x^2 + 2) $$

$$ = 2x\sqrt{3}(2x^2 + 3) $$

So, $ \frac{dy}{dx} = 2x\sqrt{3}(2x^2 + 3) $.