Finding the derivative of $y= (ln x)^2$
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In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:
$y=(ln x)^2$.
First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.
$y=frac{3}{sqrt{2x+1}}$
First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.
I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.
derivatives
asked Dec 3 '18 at 0:41
EllaElla
33111
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add a comment |
$begingroup$
In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:
$y=(ln x)^2$.
First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.
$y=frac{3}{sqrt{2x+1}}$
First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.
I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.
derivatives
asked Dec 3 '18 at 0:41
EllaElla
33111
$endgroup$
add a comment |
$begingroup$
In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:
$y=(ln x)^2$.
First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.
$y=frac{3}{sqrt{2x+1}}$
First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.
I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.
derivatives
asked Dec 3 '18 at 0:41
EllaElla
33111
$endgroup$
In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:
$y=(ln x)^2$.
First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.
$y=frac{3}{sqrt{2x+1}}$
First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.
I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.
derivatives
derivatives
asked Dec 3 '18 at 0:41
EllaElla
33111
asked Dec 3 '18 at 0:41
EllaElla
33111
edited Dec 3 '18 at 0:52
Ella
asked Dec 3 '18 at 0:41
EllaElla
33111
asked Dec 3 '18 at 0:41
EllaElla
33111
asked Dec 3 '18 at 0:41
EllaElla
33111
33111
add a comment |
add a comment |
1 Answer
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First one is correct, for the second one
begin{eqnarray}
frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
&=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
&=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
&=& -frac{3}{(2x + 1)^{3/2}}
end{eqnarray}
edited Dec 3 '18 at 1:40
sound wave
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
First one is correct, for the second one
begin{eqnarray}
frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
&=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
&=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
&=& -frac{3}{(2x + 1)^{3/2}}
end{eqnarray}
edited Dec 3 '18 at 1:40
sound wave
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
$endgroup$
add a comment |
$begingroup$
First one is correct, for the second one
begin{eqnarray}
frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
&=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
&=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
&=& -frac{3}{(2x + 1)^{3/2}}
end{eqnarray}
edited Dec 3 '18 at 1:40
sound wave
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
$endgroup$
add a comment |
$begingroup$
First one is correct, for the second one
begin{eqnarray}
frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
&=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
&=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
&=& -frac{3}{(2x + 1)^{3/2}}
end{eqnarray}
edited Dec 3 '18 at 1:40
sound wave
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
$endgroup$
First one is correct, for the second one
begin{eqnarray}
frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
&=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
&=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
&=& -frac{3}{(2x + 1)^{3/2}}
end{eqnarray}
edited Dec 3 '18 at 1:40
sound wave
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
edited Dec 3 '18 at 1:40
sound wave
1738
edited Dec 3 '18 at 1:40
sound wave
1738
edited Dec 3 '18 at 1:40
sound wave
1738
1738
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
answered Dec 3 '18 at 1:02
caveraccaverac
14.4k31130
14.4k31130
add a comment |
add a comment |
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