Help chain rule/tangent stuff

[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 f(x)=f(u(x)) then

\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}

as examples, first one

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

u(x)=4x^3-6x^2

f(u)=\frac{7}{u^3}

given that

\frac{df}{du}=-\frac{21}{u^4}

and

\frac{du}{dx}=12x^2-12x

then chain rule states

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

substituting u

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

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

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

y(x)=\frac{x}{4}(3 \pm \sqrt{5})

or by implicit differentiation.

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

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

[Roxy]

Thaaaaank you :biggrin: