Thread: [SOLVED] Diferentiation using chain rule?

1. [SOLVED] Diferentiation using chain rule?

I'm not sure how to go about doing these problems.
1. y = sin( ( 3/x^3) - (2/x^2) + (1/x) )
do i distribute sin to all the fraction and then differentite or do i use the quotient rule inside the bracket and sin x = cos x???

2. cuberoot(cos(3x)
still not sure how to go about it. if i use the chain rule i have 3 function and my teacher only explained using 2 functions... how it still work?

thank you so much for any help

2. Originally Posted by simply07
I'm not sure how to go about doing these problems.
1. y = sin( ( 3/x^3) - (2/x^2) + (1/x) )
do i distribute sin to all the fraction and then differentite or do i use the quotient rule inside the bracket and sin x = cos x???

2. cuberoot(cos(3x)
still not sure how to go about it. if i use the chain rule i have 3 function and my teacher only explained using 2 functions... how it still work?

thank you so much for any help
Sometimes depending on the question, you will need to use the chain rule more then once or twice....I've used it 7 times before.

For the first, you don't do anything to the fraction, just use chain rule.

1. $\dfrac{dy}{dx}=(cos[\dfrac{3}{x^3}-\dfrac{2}{x^2}+\dfrac{1}{x}])(\dfrac{4}{x^3}-\dfrac{9}{x^4}-\dfrac{1}{x^2})$

$\dfrac{dy}{dx}=\dfrac{1}{3}(cos(3x))^{\frac{-2}{3}}(-sin(3x)(3))$
$\dfrac{dy}{dx}=\dfrac{-sin(3x)}{(cos(3x))^{\frac{2}{3}}}$