Chain Rule - Differential Calculus I am having a Calculus test tomorrow ,and I've been practicing. I encountered this problem which I can't seem to find the solution: $
5x(x-8)^{50}$. The textbook says that the answer is $5(x-8)^{49}(51x-8)$. I would like to know the way the book did it.
Thanks in advance.
 A: $${ \left( 5x(x-8)^{ 50 } \right)  }^{ \prime  }={ \left( 5x \right)  }^{ \prime  }{ \left( x-8 \right)  }^{ 50 }+5x{ \left( { \left( x-8 \right)  }^{ 50 } \right)  }^{ \prime  }=5{ \left( x-8 \right)  }^{ 50 }+250x{ \left( x-8 \right)  }^{ 49 }={ \left( x-8 \right)  }^{ 49 }\left( 5x-40+250x \right) =\\ ={ \left( x-8 \right)  }^{ 49 }\left( 255x-40 \right) =5\left( 51x-8 \right) { \left( x-8 \right)  }^{ 49 }$$
A: I am assuming you mean to differentiate $f(x) = 5x(x-8)^{50}$ with respect to $x$. 
By the Product Rule:
$$ f'(x) = (\frac{d}{dx}(5x))\cdot (x-8)^{50} + (\frac{d}{dx}(x-8)^{50})\cdot 5x$$
By $\frac{d}{dx}(x^n) = nx^{n-1}$  :
$$ f'(x) = 5\cdot (x-8)^{50} + 50\cdot (x-8)^{49}\cdot 5x$$ 
$$ f'(x) = 5(x-8)^{49}((x-8)+50x)$$
$$ f'(x) = 5(x-8)^{49}(51x-8)$$
A: A simple substitution will make differentiating this function easier. Let $u=(x-8)$. Then $x=(u+8)$ and
$$\\ 5x(x-8)^{50}=5(u+8)u^{50}=5(u^{51}+8u^{50})\\$$
Differentiating with respect to the original variable $x$ requires using the Chain Rule (though trivial in this case as $du = 1dx$):
$$\frac{d}{dx}5x(x-8)^{50} \\$$
$$=\frac{d}{dx}5(u^{51}+8u^{50}) \\$$
$$= 5\frac{d}{du}(u^{51}+8u^{50})\frac{du}{dx}\\$$
$$= 5(51u^{50}+8\cdot50u^{49})(1) \\$$
$$= 5u^{49}(51u+8\cdot50)$$
Re-inserting our original variable,
$$=5(x-8)^{49}(51(x-8)+50\cdot8) \\$$
$$=5(x-8)^{49}(51x-51\cdot8+50\cdot8) \\$$
$$=5(x-8)^{49}(51x-8)$$
Notice the pattern of the original problem, where simple factor $x$ had a nice power of 1, and the complicated factor $(x-8)$ had an undesirable power of 50:
$$5x(x-8)^{50}=5(u+8)u^{50}$$
Notice how our substitution $u=x-8$ gave us a simple factor of $u$ beneathe the power of 50, allowing us to avoid the product rule. If you like this answer, feel free to activate the check mark next to it.
A: If you want to practice the chain rule:
$$\left[5x(x-8)^{50}\right]'=\left[5(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})^{50} \right]'=5\cdot50\cdot(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})^{49}\cdot(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})'=$$
$$250\cdot x^{\frac{49}{50}}(x-8)^{49}\cdot \left(\frac{51}{50}x^{\frac{1}{50}}-\frac{8}{50}x^{-\frac{49}{50}}\right)=250\cdot\frac{1}{50}\cdot(x-8)^{49}\cdot\left(51x^{\frac{49}{50}+\frac{1}{50}}-8x^{\frac{49}{50}-\frac{49}{50}}\right)=5(x-8)^{49}(51x-8).$$
