# finding derivatives

• Aug 26th 2009, 10:29 PM
deej813
finding derivatives
find the derivative of:

a) f(t)=(4t+5)^4

b) g(x)=(x+5)/(x-2)
• Aug 26th 2009, 10:39 PM
Chris L T521
Quote:

Originally Posted by deej813
find the derivative of:

a) f(t)=(4t+5)^4

b) g(x)=(x+5)/(x-2)

For a), apply the chain rule:

$\displaystyle \left[f(g(t))\right]^{\prime}=f^{\prime}(g(t))\cdot g^{\prime}(t)$.

In you case, let $\displaystyle f(t)=t^4$ and $\displaystyle g(t)=4t+5$.

For b), apply quotient rule:

$\displaystyle \left[\frac{f(x)}{g(x)}\right]^{\prime}=\frac{g(x)f^{\prime}(x)-f(x)g^{\prime}(x)}{\left[g(x)\right]^2}$

In your case, let $\displaystyle f(x)=x+5$ and $\displaystyle g(x)=x-2$.

Can you take it from here?
• Aug 26th 2009, 10:48 PM
deej813
umm i haven't learnt the chain rule or the quotient rule yet
is there another way to do it??
• Aug 26th 2009, 10:51 PM
Chris L T521
Quote:

Originally Posted by deej813
umm i haven't learnt the chain rule or the quotient rule yet
is there another way to do it??

Quotient rule is needed for (b), but for (a) you can multiply out $\displaystyle (4t+5)^4$ and differentiate termwise.

--OR--

You apply the limit definition of the derivative...
• Aug 26th 2009, 10:56 PM
deej813
ok well multiplying it out and then taking the derivative i get
1023t^3 +3840t^2 + 4800t +2000

the answer says 16(4t+5)^3
• Aug 26th 2009, 11:03 PM
Chris L T521
Quote:

Originally Posted by deej813
ok well multiplying it out and then taking the derivative i get
1023t^3 +3840t^2 + 4800t +2000

the answer says 16(4t+5)^3

The answer they gave was achieved by using the chain rule. I'm surprised they have the answer in that form, given that you told me that you don't know the rule.

They did it as follows:

$\displaystyle \frac{\,d}{\,dt}\left[\left(4t+5\right)^4\right]=4\left(4t+5\right)^3\cdot\frac{\,d}{\,dt}\left[4t+5\right]=4\left(4t+5\right)^3\cdot 4=16\left(4t+5\right)^3$
• Aug 26th 2009, 11:10 PM
deej813
ok thanks
would you be able to explain how that works then
• Aug 27th 2009, 11:52 AM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/diffChain/simple.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule). So, imagine the dashed balloon is just a variable, like x or u etc...

http://www.ballooncalculus.org/asy/d...in/simple2.png

... to decide the multiplier and the power in the derivative. (In this case, 4 and one less than 4, respectively.)

The second one...

http://www.ballooncalculus.org/asy/d...leQuotient.png

... where the product rule (legs crossed or uncrossed)...

http://www.ballooncalculus.org/asy/prod.png

shows the pattern for differentiating a product - in this case, x + 5 times the fraction 1/(x - 2), but we write the fraction as (x - 2) to the power minus one, and apply the chain rule again, to perform the right hand part of the (legs uncrossed) product rule.

Tweak the second balloon on the bottom row...

http://www.ballooncalculus.org/asy/d...eQuotient2.png

... so that a suitable common denominator is ready (for simplifying the bottom row) and you can see how the quotient rule is derived from the other two (so you never really need it... and if I've made the first two rules seem really scary, then at least they're all you need!)

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