What is the derivative of # f(x) = x^5 * (x^2-3)^6#?

1 Answer
Mar 7, 2018

Answer:

#5x^4(x^2-3)^6+12x^6(x^2-3)^5#

Explanation:

Here:

#d/dx x^5 * (x^2-3)^6#

we can use product rule:

#d/dx color(red)a * color(blue)b = (color(red)(a))'(color(blue)b) + (color(red)a)(color(blue)b)'#

So:

#d/dx color(red)(x^5) * color(blue)((x^2-3)^6)#

becomes:

#(color(red)x^5)'color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'#

Simplifying:

#(5x^4)color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'#

#d/dx color(blue)((x^2-3)^6)#

We can use chain rule here:

#d/dxf(x) = d/(du)f(u) * d/dx (x)#

#->d/dx(x^2-3)^6#

becomes:

#d/dx (u)^6 * d/dx (x^2-3)#

#=6u^5*2x#

Since #u=(x^2-3)#:

#=6(x^2-3)^5*2x#

#d/dx color(blue)((x^2-3)^6)=12x(x^2-3)^5#

Simplifying our former equation:

#(5x^4)color(blue)((x^2-3)^6)+(color(red)x^5)(color(blue)((x^2-3)^6))'#

becomes:

#(5x^4)(x^2-3)^6+(x^5)(12x)(x^2-3)^5#

Multiplying it out:

#=5x^4(x^2-3)^6+12x^6(x^2-3)^5#

And there we have our answer