What is the Chain Rule and How Does it Apply to Tangent Functions?

[Roxy]

I was trying it but i got stuck.

http://img56.exs.cx/my.php?loc=img56&image=scan7xv.jpg

http://img28.exs.cx/my.php?loc=img28&image=scan30fn.jpg

 

[ReyChiquito]

chain rule states that if [itex]f(x)=f(u(x))[/itex] then

[tex]\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}[/tex]

as examples, first one

[tex]f(x)=\frac{7}{(4x^3-6x^2)^3}[/tex]

[tex]u(x)=4x^3-6x^2[/tex]

[tex]f(u)=\frac{7}{u^3}[/tex]

given that

[tex]\frac{df}{du}=-\frac{21}{u^4}[/tex]

and

[tex]\frac{du}{dx}=12x^2-12x[/tex]

then chain rule states

[tex]\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=-\frac{21}{u^4}(12x^2-12x)[/tex]

substituting [itex]u[/itex]

[tex]\frac{df}{dx}=-\frac{252}{(4x^3-6x^2)^4}(x^2-x)[/tex]

in the next one, you are almost there... all you have to do is factorize [itex](6x^2-5)^4(2x-1)^3[/itex]

the third one you can do it by calculating [itex]y=y(x)[/itex] and [itex]x=x(y)[/itex] and then differentiate ie

[tex]y(x)=\frac{x}{4}(3 \pm \sqrt{5})[/tex]

or by implicit differentiation.

The next one you already have it, the only thing you need to do is evaluate [itex]y'(2)[/itex] and [itex]y'(3)[/itex] then use the fact that
[tex]m=\frac{f'(3)-f'(2)}{3-2}[/tex]

in the last two you use the fact that [itex](f\circ g)(x)=g(f(x))[/itex] be carefull tough, you are doing it wrong in the last one (watch the root).

 

[Roxy]

Thaaaaank you

Thaaaaank you