Question

Use the chain rule to find the derivative.

Answer 1

\[f(x)=(2x-3)^4(x^2+x+1)^5\]

Answer 2

so id use the product rule first..

Answer 3

I think yo u would use the product rule first if im not too rusty

Answer 4

\[(2x-3)^4 5(x^2+x+1)^4 + (x^2 + x +1)^54(2x-3)^3\]

Answer 5

then, the chain rule f'(g(x))(g'(x))

Answer 6

\[\frac{d((x^2+x+1)^5)}{dx} \ne 5(x^2+x+1)^5\]

Answer 7

With the product rule, jsut leave the one you differentiated in the form on the left- you started the chain rule prematurely.

Answer 8

ok so

Answer 9

\[(2x-3)^4(x^2+x+1)^5' + (x^2+x + 1)^5(2x-3)^4\]

Answer 10

Yes, that's correct.

Answer 11

ok now for chain rule.. this is where I get lost big time

Answer 12

OK\[\frac{d(x^2+x+1)^5}{dx}=\frac{d(x^2+x+1)^5}{d(x^2+x+1)} \times \frac{d(x^2+x+1)}{dx}\]

Answer 13

For the first one, just treat (x^2+x+1) as a single variable

Answer 14

Call it a or whatever:\[\frac{da^5}{da}=5a^4=5(x^2+x+1)^4\]

Answer 15

i just dont know how to set it up, i know thats the derivative of that

Answer 16

with my product rule like that, where do i start?

Answer 17

Differentiate the bits that need to be differentiated. What do you find hard about this problem?

Answer 18

ok here I go

Answer 19

(2x-3)^4 5(x^2+x+1)^4 + (x^2 + x + 1)^5 4(2x-3)^3

Answer 20

thats from the product rule

Answer 21

No, It's not.

Answer 22

I'll go through the problem in full.

Answer 23

ok can you explain it as if i were a five year old?

Answer 24

\[\frac{d(2x-3)^4 (x^2+x+1)^5}{dx}=\frac{d(2x-3)^4 }{dx}(x^2+x+1)^5+\frac{d (x^2+x+1)^5}{dx}(2x-3)^4\]\[\frac{d(2x-3)^4 }{dx}=\frac{d(2x-3)^4 }{d(2x-3)}\frac{d(2x-3)}{dx}\]\[q=(2x-3)\]
\[\frac{d(2x-3)^4 }{dx}=\frac{d(q)^4 }{d(q)} \times \frac{d(2x-3)}{dx}=4q^3 \times 2\]
Substitute for q.

Answer 25

\[\frac{d(a \times b)}{dx}=\frac{d(a)}{dx} \times b+\frac{d(b)}{dx} \times a\]

Answer 26

im losing you

Answer 27

That's the product rule

By the way, it's better to write derivitaves as d?/dx than ?'

Answer 28

Do you understand the first line of the proof?

Answer 29

yes i have the product rule down

Answer 30

f(g)' + g(f)'

Answer 31

f= (2x-3)^4 and g = (x^2+x+1)

Answer 32

^5

Answer 33

(I'm not the best at explaining things, but http://djm.cc/library/Calculus_Made_Easy_Thompson.pdf page 49 (product rule) and 81 (chain rule) are very good)

Answer 34

As I said, it REALLY confuses things writing down (f)'

Write\[\frac{df}{dx}\]

Answer 35

\[\frac{dfg}{dx}=\frac{df}{dx}g+\frac{dg}{dx}f\]

Answer 36

yep thats product rule

Answer 37

That's what my first line says.

Answer 38

The second line goes into HOW to work out\[\frac{df}{dx}\]

Answer 39

f(g)' = (2x-3)^4 5(x^2+x+1)^4

Answer 40

You're overcomplicating. Simply work out d(2x-3)^2/dx and d(x^2-x+1)^2/dx, and substitute into the first line.

Answer 41

Do you know how to work out d(2x-3)^2/dx, for example?

Answer 42

yes it would be 2(2x-3)(2x-3)'

Answer 43

I've got to go. We're probably in different time zones, but when I get back I'll answer your questions in full.

Answer 44

Again, don't use(2x-3)', really confusing. (2x-3)'=2, anyway

Answer 45

So the whole thing=4(2x-3)

Answer 46

yep

Answer 47

i think i found a stupid but right way of doing it

Answer 48

so from the product rule
(x^2+x+1)^5 4(2x-3)^3 (2) + (2x+3)^4 5(x^2+x+1) (2x+1)

Answer 49

= 8(2x-3)^3(x^2+x+1)^5 + (2x+3)^4 5(x^2+x +1)(2x+1)