derivative
- Thread starter oded244
- Start date
#1: product rule. Rewrite as \(\displaystyle \L\\\underbrace{x^{4}}_{\text{f(x)}}\overbrace{(x+1)^{-4}}^{\text{g(x)}}\)
\(\displaystyle \L\\f(x)g'(x)+g(x)f'(x)\)
\(\displaystyle \L\\\underbrace{x^{4}}_{\text{f(x)}}\overbrace{(-4)(x+1)^{-5}}^{\text{g'(x)}}+\underbrace{(x+1)^{-4}}_{\text{g(x)}}\overbrace{(4x^{3})}^{\text{f'(x)}}\)
Now simplify.
#2: Same as above. product rule.
\(\displaystyle \L\\f(x)g'(x)+g(x)f'(x)\)
\(\displaystyle \L\\\underbrace{x^{4}}_{\text{f(x)}}\overbrace{(-4)(x+1)^{-5}}^{\text{g'(x)}}+\underbrace{(x+1)^{-4}}_{\text{g(x)}}\overbrace{(4x^{3})}^{\text{f'(x)}}\)
Now simplify.
#2: Same as above. product rule.
Since you understand the product rule, you must use a little chain rule on it also. Remember, derivative of inside times derivative of outside.
The second part is just:
Derivative of outside: \(\displaystyle \L\\3(5x-1)^{2}\)
Derivative of inside is just 5
So, you have: \(\displaystyle \L\\3(5x-1)^{2}(5)=15(5x-1)^{2}\)
Now, combine it with the other steps and you have it.
The second part is just:
Derivative of outside: \(\displaystyle \L\\3(5x-1)^{2}\)
Derivative of inside is just 5
So, you have: \(\displaystyle \L\\3(5x-1)^{2}(5)=15(5x-1)^{2}\)
Now, combine it with the other steps and you have it.
