Finding the derivative of $y= (ln x)^2$
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
Ella
32111
add a comment |
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
Ella
32111
add a comment |
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
Ella
32111
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
Ella
32111
asked Dec 3 '18 at 0:41
Ella
32111
edited Dec 3 '18 at 0:52
asked Dec 3 '18 at 0:41
Ella
32111
asked Dec 3 '18 at 0:41
Ella
32111
asked Dec 3 '18 at 0:41
Ella
32111
32111
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
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
add a comment |
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1 Answer
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1 Answer
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active
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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
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
add a comment |
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
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
add a comment |
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
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
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
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
edited Dec 3 '18 at 1:40
sound wave
1618
edited Dec 3 '18 at 1:40
sound wave
1618
edited Dec 3 '18 at 1:40
sound wave
1618
1618
answered Dec 3 '18 at 1:02
caverac
13.8k21130
answered Dec 3 '18 at 1:02
caverac
13.8k21130
answered Dec 3 '18 at 1:02
caverac
13.8k21130
13.8k21130
add a comment |
add a comment |
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