Differentiate each function. Do not expand an expression before differentiating. a)f(x)=(2x+3)^4

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Differentiate each function. Do not expand an expression before differentiating. a)f(x)=(2x+3)^4

Mathematics
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f(x)=4(2x+3)^3
is it be like that
use chain rule

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|dw:1443580871976:dw|
where u(x)=2x+3 and n=4
oh ok f(x)= 4(2x+3)^3 (2) = 8(2x+3)^3
good work
thnx
g(x)=(x^2-4)^3 =3(x^2-4)^2 =6x(x^2-4)^2 For this question Is it be like this!!
ayeee what tower is that in the back, lol why this place look so familiar!
lol its nagara falls!!!
hahahaha
thats skylone right
nah clearly that looks like the sydney harbour bridge
lmao pls xD
yes you are right @MTALHAHASSAN2 with the other question
ok thnx again
you should not write g(x)=(x^2-4)^3 =3(x^2-4)^2 =6x(x^2-4)^2 as none of those are equal to each other
how about this one: f(x)= (pie^2-x^2)3
3(pie^2-x^2)2
3(pie^2-x^2)^2 (2pie-x)
\[\large\rm f(x)= (\pi^2-x^2)^3\]Your first step looks good, applying power rule.\[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{royalblue}{(\pi^2-x^2)'}\]But you also need chain rule.
Ah very close!
Derivative of x^2 is not x. And derivative of pi^2 is not 2pi.
i know its 2x
2pi is just a number. It's a fancy looking constant. Remember how to differentiate a constant? \(\large\rm \frac{d}{dx}3=?\)
pi^2 is just a number* blah typo :)
then is it be 6(pie^2-x^2)^2
@zepdrix is it right??
but at the back of the book they have -6 instead of 6
oh sorry i ran off for a sec >.<
oh ok np
\[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{royalblue}{(\pi^2-x^2)'}\]\[\large\rm f'(x)= 3(\pi^2-x^2)^2\color{orangered}{(0-2x)}\]Ya, don't forget about the negative in front of the x^2
\[\large\rm f'(x)= 3(\pi^2-x^2)^2(-2x)\]\[\large\rm f'(x)= -6x(\pi^2-x^2)^2\]Understand why there is a negative? :o
yeah got you and thnx a lot

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