# chain rule

• Jun 15th 2006, 05:45 PM
c_323_h
chain rule
I need some help to differentiate

$\displaystyle g(x)=(1+4x)^5(3+x-x^2)^8$
• Jun 15th 2006, 06:00 PM
malaygoel
Quote:

Originally Posted by c_323_h
I need some help to differentiate

$\displaystyle g(x)=(1+4x)^5(3+x-x^2)^8$

Let (1+4x)^5 =f(x)
and (3+x-x^2)=h(x)
Differentiation of g(x) = differentiation of f(x)h(x)
= [diff. of f(x)]*h(x) + [diff. of h(x)]*f(x)
diff. of f(x) = 5[(1+4x)^4]*4
diff. of h(x)= 8(3+x-x^2)^7{1-2x}
• Jun 15th 2006, 09:27 PM
Soroban
Hello, c_323_h!

Quote:

Differentiate: $\displaystyle g(x) \:=\:(1+4x)^5(3+x-x^2)^8$

Product Rule: .$\displaystyle g'(x)\;=\;(1 + 4x)^5\cdot\frac{d}{dx}(3 + x - x^2)^8 +$ $\displaystyle (3 + x - x^2)^8\cdot\frac{d}{dx}(1 + 4x)^5$

Chain Rule: .$\displaystyle g'(x)\;=\;(1 + 4x)^5\cdot8(3 + x - x^2)^7(1 - 2x) + (3 + x - x^2)^8$$\displaystyle \cdot5(1 + 4x)^4\cdot4$

Factor: .$\displaystyle g'(x)\;=\;4(1 + 4x)^4(3 + x - x^2)^7$ $\displaystyle [2(1+4x)(1-2x) + 5(3+x-x^2)]$

Simplify: .$\displaystyle g'(x)\;=\;4(1+4x)^4(3+x-x^2)^7(17+9x-21x^2)$