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How do I evaluate the following derivative?
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How can I evaluate the derivative
$$left. fracddx left( x , ln(x^5) right)right|_x=1 , ?$$
I have this math problem, and the biggest thing I do not understand, is which derivative rule do I need to apply, and what does the vertical bar mean in this problem?
Solution
$$left. fracddx left( x , ln(x^5) right)right|_x=1 = 5.$$
calculus derivatives
edited 1 hour ago
Leucippus
19k92769
asked 2 hours ago
Jeune
112
add a comment |Â
up vote
2
down vote
favorite
How can I evaluate the derivative
$$left. fracddx left( x , ln(x^5) right)right|_x=1 , ?$$
I have this math problem, and the biggest thing I do not understand, is which derivative rule do I need to apply, and what does the vertical bar mean in this problem?
Solution
$$left. fracddx left( x , ln(x^5) right)right|_x=1 = 5.$$
calculus derivatives
edited 1 hour ago
Leucippus
19k92769
asked 2 hours ago
Jeune
112
The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
1
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How can I evaluate the derivative
$$left. fracddx left( x , ln(x^5) right)right|_x=1 , ?$$
I have this math problem, and the biggest thing I do not understand, is which derivative rule do I need to apply, and what does the vertical bar mean in this problem?
Solution
$$left. fracddx left( x , ln(x^5) right)right|_x=1 = 5.$$
calculus derivatives
edited 1 hour ago
Leucippus
19k92769
asked 2 hours ago
Jeune
112
How can I evaluate the derivative
$$left. fracddx left( x , ln(x^5) right)right|_x=1 , ?$$
I have this math problem, and the biggest thing I do not understand, is which derivative rule do I need to apply, and what does the vertical bar mean in this problem?
Solution
$$left. fracddx left( x , ln(x^5) right)right|_x=1 = 5.$$
calculus derivatives
calculus derivatives
edited 1 hour ago
Leucippus
19k92769
asked 2 hours ago
Jeune
112
edited 1 hour ago
Leucippus
19k92769
asked 2 hours ago
Jeune
112
edited 1 hour ago
Leucippus
19k92769
edited 1 hour ago
Leucippus
19k92769
edited 1 hour ago
Leucippus
19k92769
19k92769
asked 2 hours ago
Jeune
112
asked 2 hours ago
Jeune
112
asked 2 hours ago
Jeune
112
112
The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
1
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago
add a comment |Â
The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
1
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago
The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
1
1
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
As an alternative since $xln(x^5)=5xln(x)$ by product rule
$$fracddx(xln(x^5))=5fracddx(xln x)=5left(1cdot ln x+xcdot frac1xright)=5ln x+5$$
answered 1 hour ago
gimusi
74.1k73889
add a comment |Â
up vote
3
down vote
$$fracddx(xln(x^5))=$$
$$ln(x^5)fracddx(x)+xfracddx(5ln(x))=$$
$$ln(x^5)+xfrac5x=$$
$$ln(x^5)+5$$
for $x=1$, it gives $0+5=5$.
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
As an alternative since $xln(x^5)=5xln(x)$ by product rule
$$fracddx(xln(x^5))=5fracddx(xln x)=5left(1cdot ln x+xcdot frac1xright)=5ln x+5$$
answered 1 hour ago
gimusi
74.1k73889
add a comment |Â
up vote
4
down vote
As an alternative since $xln(x^5)=5xln(x)$ by product rule
$$fracddx(xln(x^5))=5fracddx(xln x)=5left(1cdot ln x+xcdot frac1xright)=5ln x+5$$
answered 1 hour ago
gimusi
74.1k73889
add a comment |Â
up vote
4
down vote
up vote
4
down vote
As an alternative since $xln(x^5)=5xln(x)$ by product rule
$$fracddx(xln(x^5))=5fracddx(xln x)=5left(1cdot ln x+xcdot frac1xright)=5ln x+5$$
answered 1 hour ago
gimusi
74.1k73889
As an alternative since $xln(x^5)=5xln(x)$ by product rule
$$fracddx(xln(x^5))=5fracddx(xln x)=5left(1cdot ln x+xcdot frac1xright)=5ln x+5$$
answered 1 hour ago
gimusi
74.1k73889
answered 1 hour ago
gimusi
74.1k73889
answered 1 hour ago
gimusi
74.1k73889
answered 1 hour ago
gimusi
74.1k73889
74.1k73889
add a comment |Â
add a comment |Â
up vote
3
down vote
$$fracddx(xln(x^5))=$$
$$ln(x^5)fracddx(x)+xfracddx(5ln(x))=$$
$$ln(x^5)+xfrac5x=$$
$$ln(x^5)+5$$
for $x=1$, it gives $0+5=5$.
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
add a comment |Â
up vote
3
down vote
$$fracddx(xln(x^5))=$$
$$ln(x^5)fracddx(x)+xfracddx(5ln(x))=$$
$$ln(x^5)+xfrac5x=$$
$$ln(x^5)+5$$
for $x=1$, it gives $0+5=5$.
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$fracddx(xln(x^5))=$$
$$ln(x^5)fracddx(x)+xfracddx(5ln(x))=$$
$$ln(x^5)+xfrac5x=$$
$$ln(x^5)+5$$
for $x=1$, it gives $0+5=5$.
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
$$fracddx(xln(x^5))=$$
$$ln(x^5)fracddx(x)+xfracddx(5ln(x))=$$
$$ln(x^5)+xfrac5x=$$
$$ln(x^5)+5$$
for $x=1$, it gives $0+5=5$.
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
answered 2 hours ago
Salahamam_ Fatima
34.2k21430
34.2k21430
add a comment |Â
add a comment |Â
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The vertical bar means to evaluate the derivative at $x=5$, i.e., find $f'(5)$ where $f(x)=xln(x^5)$.
– Barry Cipra
2 hours ago
The vertical bar with $x=1$ means you should take the derivative at $x=1$.
– Torsten Schoeneberg
2 hours ago
So all I need to do, is just plug in 5 in for x? Do I take the derivative before or after plugging in x=5?
– Jeune
2 hours ago
1
Oh, sorry, I meant to say $x=1$, not $x=5$, so it's $f'(1)$. You take the derivative of $f(x)=xln(x^5)$ first, then plug in $x=1$ (not $x=5$).
– Barry Cipra
2 hours ago
Welcome to math.SE!! In this site we use a program called MathJax to write the math correctly. Please consider it so we all speak the same language.
– manooooh
1 hour ago