How do you differentiate #(3x^2+4) / (sqrt(1+x^2))#?

Hey there! :)

Looks tough when you first see it, right? However, differentiation rules are like super powers.

Here is the short story:

#d/dx((3x^2+4)/sqrt(1+x^2))= (6x)/sqrt(1+x^2)-(x(3x^2+4))/(1+x^2)^(3/2)#

Now for the far more exciting long story:

The quotient rule says that if we have two differentiable functions, say #f(x) and g(x)# such that #g(x)!=0#, then we have

#d/dx(f(x)/g(x)) = (d/dx(f(x))*g(x)-f(x)*d/dx(g(x)))/(g(x))^2#

So now we just apply this!

So,

#d/dx((3x^2+4)/sqrt(1+x^2))=(d/dx(3x^2+4)*sqrt(1+x^2)-(3x^2+4)*d/dx(sqrt(1+x^2)))/(sqrt(1+x^2))^2#

Now we have to use a few super powers here. The power rule and chain rule are part of our arsenal.

I hope it is clear that

#d/dx(3x^2+4) = 2*3x + 0=6x#

(If not, just shout!) - We just applied the power rule to the first term and the second term is a constant. And constants don't stand a chance against differentiation! They just become zero.

Next, we harness the power of the chain rule, that says for any differentiable functions, say, #u(x)# and #h(x)#,

#d/dx(u(h(x)))=d/(dh(x))(u(h(x)))*d/dx(h(x))#

We apply it (here u is the square root function and h(x) = (1+x^2)):

#d/dx(sqrt(1+x^2))=1/2*(1+x^2)^(-1/2)*(2x) = x(1+x^2)^(-1/2)#

Finally, plugging these things back in their rightful place, we get

# d/dx((3x^2+4)/sqrt(1+x^2))=(6x*sqrt(1+x^2)-(3x^2+4)*x(1+x^2)^(-1/2))/(1+x^2) #
#=(6xsqrt(1+x^2))/(1+x^2)-(x(3x^2+4))/((1+x^2)*sqrt(1+x^2))#
#=(6x)/sqrt(1+x^2)-(x(3x^2+4))/((1+x^2)^(3/2))#