Question
Function
f′(x)=(x−3)22ax3−15ax2+36ax−24a
Evaluate
f(x)=(x−2)×x−3x−4×xa
Simplify
More Steps

Evaluate
(x−2)×x−3x−4×xa
Multiply the terms
More Steps

Evaluate
x−3x−4×xa
Multiply the terms
x−3x(x−4)×a
Multiply the terms
x−3x(x−4)a
(x−2)×x−3x(x−4)a
Multiply the terms
x−3(x−2)x(x−4)a
f(x)=x−3(x−2)x(x−4)a
Evaluate
f(x)=x−3ax(x−2)(x−4)
Take the derivative of both sides
f′(x)=dxd(x−3ax(x−2)(x−4))
Calculate
f′(x)=dxd(x−3ax3−6ax2+8ax)
Use differentiation rule dxd(g(x)f(x))=(g(x))2dxd(f(x))×g(x)−f(x)×dxd(g(x))
f′(x)=(x−3)2dxd(ax3−6ax2+8ax)×(x−3)−(ax3−6ax2+8ax)×dxd(x−3)
Calculate
More Steps

Evaluate
dxd(ax3−6ax2+8ax)
Use differentiation rule dxd(f(x)±g(x))=dxd(f(x))±dxd(g(x))
dxd(ax3)+dxd(−6ax2)+dxd(8ax)
Calculate
3ax2+dxd(−6ax2)+dxd(8ax)
Calculate
3ax2−12ax+dxd(8ax)
Calculate
3ax2−12ax+8a
f′(x)=(x−3)2(3ax2−12ax+8a)(x−3)−(ax3−6ax2+8ax)×dxd(x−3)
Calculate
More Steps

Evaluate
dxd(x−3)
Use differentiation rule dxd(f(x)±g(x))=dxd(f(x))±dxd(g(x))
dxd(x)−dxd(3)
Use dxdxn=nxn−1 to find derivative
1−dxd(3)
Use dxd(c)=0 to find derivative
1−0
Removing 0 doesn't change the value,so remove it from the expression
1
f′(x)=(x−3)2(3ax2−12ax+8a)(x−3)−(ax3−6ax2+8ax)×1
Any expression multiplied by 1 remains the same
f′(x)=(x−3)2(3ax2−12ax+8a)(x−3)−(ax3−6ax2+8ax)
Solution
More Steps

Evaluate
(3ax2−12ax+8a)(x−3)−(ax3−6ax2+8ax)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
(3ax2−12ax+8a)(x−3)−ax3+6ax2−8ax
Expand the expression
3ax3−21ax2+44ax−24a−ax3+6ax2−8ax
Subtract the terms
2ax3−21ax2+44ax−24a+6ax2−8ax
Add the terms
2ax3−15ax2+44ax−24a−8ax
Subtract the terms
2ax3−15ax2+36ax−24a
f′(x)=(x−3)22ax3−15ax2+36ax−24a
Show Solution
