## mlddmlnog Group Title 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)=? one year ago one year ago

1. mlddmlnog Group Title

2. mlddmlnog Group Title

@zepdrix

3. zepdrix Group Title

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

4. zepdrix Group Title

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

5. mlddmlnog Group Title

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

6. mlddmlnog Group Title

yay i got it :)

7. zepdrix Group Title

yay

8. mlddmlnog Group Title

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 Group Title

Hmmmmm....

10. mlddmlnog Group Title

11. zepdrix Group Title

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 Group Title

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 Group Title

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 Group Title

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.