Are these two equivalent
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When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
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add a comment |
$begingroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
$endgroup$
add a comment |
$begingroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
$endgroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
calculus
asked Dec 10 '18 at 7:33
mingming
3415
3415
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add a comment |
2 Answers
2
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They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
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$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
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– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
$endgroup$
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
$endgroup$
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
$endgroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
74.9k42865
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
$endgroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
45.7k1848
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
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