How do you differentiate #z=t^2/((t-4)(2-t^3))# using the quotient rule?

Question

Scarlett Allison · Accepted Answer

#=(2t^5-4t^4+2t^2-16t)/((2-t^3)^2 * (t-4)^2)#

Quotient Rule states that:

#d/dx (color(red)a/color(blue)b)=((color(red)a)' * (color(blue)b)-(color(blue)b)' * (color(red)a))/color(blue)b^2#

Here, we can substitute:

#d/dx ((color(red)(t^2))' * ((color(blue)(t-4))(color(blue)(2-t^3))- ((color(blue)(t-4)) * (color(blue)(2-t^3)))' * (color(red)(t^2)))) / (((color(blue)(2-t^3)) * (color(blue)(t-4)))^2#

Derivative of #color(red)(t^2) = color(red)(2t)#

Derivative of #color(blue)((t-4)(2-t^3))# is a little complicated. We must use Product Rule , which states:

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

We can substitute:

#-> (color(red)(t-4))' * (color (blue)(2-t^3)) + (color(red)(t-4)) * (color(blue)(2-t^3))'#

#= color(red)1 * (color(blue)(2-t^3)) + (color(red)(t-4)) * (color(blue)(3t^2))#

#=color(blue)(2-t^3+3t^3-12t^2)#

#d/dx (color(blue)(t-4)) * (color(blue)(2-t^3))=color(blue)(2t^3-12t^2+2)#

Now, we can substitute everything into our first equation:

#-> ((color(red)(2t)) * (color(blue)(t-4))(color(blue)(2-t^3)) - (color(blue)(2t^3-12t^2+2)) * (color(red)(t^2)))/(((color(blue)(2-t^3)) * (color(blue)(t-4)))^2#

Simplifying:

#((color(red)(−2t^5+8t^4+4t^2−16t)) - (color(blue)(2t^5−12t^4+2t^2)))/((color(blue)(2-t^3))^2(color(blue)(t-4))^2#

#=(2t^5-4t^4+2t^2-16t)/((2-t^3)^2 * (t-4)^2)#

Thus, solved.

William Abdalla · Answer

undefined To differentiate $ z = \frac{t^2}{(t-4)(2-t^3)} $ using the quotient rule, first identify $ u = t^2 $ and $ v = (t-4)(2-t^3) $. Then, apply the quotient rule formula:

$$ z' = \frac{u'v - uv'}{v^2} $$

$$ u' = 2t $$ and $ v' = -\frac{3t^2 - 8}{(2 - t^3)^2} $.

Substitute into the formula and simplify to find the derivative.