a)
[tex]f(x)=2kx^4-2x^3+3x-7,\qquad\quad k=18\\\\\\f(x)=36x^4-2x^3+3x-7\\\\ f'(x)=(36x^4-2x^3+3x-7)'=36\cdot4x^3-3\cdot3x^2+3-0=144x^3-9x^2+3[/tex]
b)
[tex]f(x)=\dfrac{x+k}{x-7}\,,\qquad\qquad k=18\\\\\\f(x)=\dfrac{x+18}{x-7}\\\\f'(x)=\left(\dfrac{x+18}{x-7}\right)'=\dfrac{(x+18)'(x-7)-(x+18)(x-7)'}{(x-7)^2}\\\\f'(x)=\dfrac{1\cdot(x-7)-(x+18)\cdot1}{(x-7)^2}=\dfrac{x-7-x-18}{(x-7)^2}=-\dfrac{25}{(x-7)^2}[/tex]
c)
We wzorze funkcji nie ma k, dlatego je zignoruję.
[tex]f(x)=x^2(3x-5)=3x^3-5x^2\\\\ f'(x)=(3x^3-5x^2)'=3\cdot3x^2-5\cdot2x=9x^2-10x[/tex]
