Confused about the difference between these two derivatives
So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.
But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??
I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?
calculus
edited Dec 10 '18 at 20:00
Bernard
118k639112
asked Dec 10 '18 at 19:57
ming
3165
add a comment |
So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.
But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??
I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?
calculus
edited Dec 10 '18 at 20:00
Bernard
118k639112
asked Dec 10 '18 at 19:57
ming
3165
1
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
It's 1, correct?
– ming
Dec 10 '18 at 20:33
add a comment |
So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.
But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??
I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?
calculus
edited Dec 10 '18 at 20:00
Bernard
118k639112
asked Dec 10 '18 at 19:57
ming
3165
So for a function $f(x) = e^{sin(x^2)}$, you're supposed to apply the chain rule to it to get $e^{sin(x^2)}(cos(x^2))(2x)$.
But for a function $f(x) = 5^{cos(x)}$, you apply the $a^x(ln(a))$ rule, to get $5^{cos(x)}ln(5)(-sin(x))$??
I'm confused, cause $5$ and $e$ are both numbers with a trig function as its exponent, why do we treat them differently?
calculus
calculus
edited Dec 10 '18 at 20:00
Bernard
118k639112
asked Dec 10 '18 at 19:57
ming
3165
edited Dec 10 '18 at 20:00
Bernard
118k639112
asked Dec 10 '18 at 19:57
ming
3165
edited Dec 10 '18 at 20:00
Bernard
118k639112
edited Dec 10 '18 at 20:00
Bernard
118k639112
edited Dec 10 '18 at 20:00
Bernard
118k639112
118k639112
asked Dec 10 '18 at 19:57
ming
3165
asked Dec 10 '18 at 19:57
ming
3165
asked Dec 10 '18 at 19:57
ming
3165
3165
1
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
It's 1, correct?
– ming
Dec 10 '18 at 20:33
add a comment |
1
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
It's 1, correct?
– ming
Dec 10 '18 at 20:33
1
1
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
It's 1, correct?
– ming
Dec 10 '18 at 20:33
It's 1, correct?
– ming
Dec 10 '18 at 20:33
add a comment |
2 Answers
2
active
oldest
votes
You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
add a comment |
For $a>0$ and $xin Bbb R$,
$$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$
the derivative is
$$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
add a comment |
You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
add a comment |
You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
You don't treat them differently. In fact, you have
$$ frac{mathrm d}{mathrm dx} a^x = a^xln(a),$$
for any $a > 0.$ In particular, when $a=e$, you land back in the usual derivative of the exponential:
$$ frac{mathrm d}{mathrm dx} e^x = e^xln(e) = e^x,$$
since $ln (e) = 1$.
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
edited Dec 10 '18 at 20:06
Chickenmancer
3,319723
3,319723
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
answered Dec 10 '18 at 19:59
MisterRiemann
5,7841624
5,7841624
add a comment |
add a comment |
For $a>0$ and $xin Bbb R$,
$$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$
the derivative is
$$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
add a comment |
For $a>0$ and $xin Bbb R$,
$$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$
the derivative is
$$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
add a comment |
For $a>0$ and $xin Bbb R$,
$$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$
the derivative is
$$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
For $a>0$ and $xin Bbb R$,
$$5^{cos(x)}=e^{cos(x)ln(5)}=e^{u(x)}$$
the derivative is
$$u'(x)e^{u(x)}=-ln(5)sin(x)e^{cos(x)ln(5)}$$
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
answered Dec 10 '18 at 20:02
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
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1
Do you know what $ln(e)$ is?
– T. Bongers
Dec 10 '18 at 19:59
It's 1, correct?
– ming
Dec 10 '18 at 20:33