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 $f(x)=f(u(x))$ 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

Thaaaaank you