mlddmlnog 2 years ago let f and g be differentiable functions such that f(1)=2 f'(1)=3 f'(2)=-4 g(1)=2 g'(1)=-3 g'(2)=5 if h(x)=f(g(x)), then h'(1)=?

1. mlddmlnog

2. mlddmlnog

@zepdrix

3. zepdrix

So you have to kind of remember the DEFINITION of the chain rule for this one.

4. zepdrix

$\large h(x)=f(g(x))$The chain rule will produce this.$\large h'(x)=f'(g(x))\cdot g'(x)$

5. mlddmlnog

oh.. wait. i'll try to do it myself now. thank you for giving me a start! :)

6. mlddmlnog

yay i got it :)

7. zepdrix

yay

8. mlddmlnog

here i the next one that's kind of similar to this. but i don't get. If $f(x)=x ^{3}+3x ^{2}+4x+5$ and g(x)=5, then g(f(x))=?

9. zepdrix

Hmmmmm....

10. mlddmlnog

11. zepdrix

So here is how a composition of functions works. Everywhere you see an x, you replace it with f(x). Example:$\large \color{orangered}{f(x)=2x}$$\large g(\color{cornflowerblue}{x})=\color{cornflowerblue}{x}+3$ $\large g(\color{orangered}{f(x)})=\color{orangered}{f(x)}+3$

12. zepdrix

In the problem we've been given, the function $$g(x)$$ is CONSTANT. There are no x's! So when we plug f(x) into it, it should give us the same answer, because g(x) is always 5. Always constant. You could do the composition thing I explained earlier and it might make sense. If you try to plug in f(x) for any x's in g, you'll see that you have nowhere to actually plug it in.

13. zepdrix

Is your teacher any good? Because so far I'm really really disliking these problems. None of them are straight forward. It just feels like he gave you a list of puzzles to work on.

14. mlddmlnog

haha yes. she loves these types of questions. -_____- all of her questions are like this. making our brains explode. it literally takes me like 8 hours to do hw. it's insane.