So it looks like you're using the limit definition of the derivative to find #d/(dx) (2x^5 - 5x^3)#. We know this will be #10x^4 - 15x^2# so it's a nice sense check when we use limits.

We're going to need to expand the brackets, in this case we'll use the binomial theorem.

#(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k#

where #((n),(k)) = (n!)/(k!(n-k)!)#

#(x+h)^5 = sum_(k=0)^5 ((5),(k)) x^(5-k)y^k#

#= ((5),(0))x^5h^0 + ((5),(1))x^4h^1 + ((5),(2)) x^3h^2 + ((5),(3))x^2h^3
+ ((5),(4))x^1h^4 + ((5),(5))x^0h^5#

# = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5#

Similarly;

#(x+h)^3 = ((3),(0))x^3h^0 + ((3),(1))x^2h + ((3),(2))xh^2 + ((3),(3))x^0h^3#

#(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3#

Collating all this into the initial function we get:

#lim_(h->0) 1/h*(2[x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5] - 5[x^3 + 3x^2h + 3xh^2 + h^3] -2x^5 + 5x^3)#

Some terms cancel out

#lim_(h->0) 1/h*(color(red)(2x^5)+10x^4h+20x^3h^2+20x^2h^3+10xh^4+2h^5 color(blue)(- 5x^3) -15x^2h - 15xh^2 - 5h^3color(red)( - 2x^5) + color(blue)(5x^3))#

Now multiplying through by #1/h#

#lim_(h->0) (10x^4+20x^3h+20x^2h^2+10xh^3+2h^4 -15x^2 - 15xh)#

As #h->0# any term with h in it will go to zero and would you look at that! It's only gone and left us with

#10x^4 - 15x^2#

Pleasingly, this is what we expected so we can be fairly confident there hasn't been any errors in the calculation.