Derivative by definition of $ f(x)= 2xsin(2x)$
$begingroup$
I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}
From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?
derivatives
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
$endgroup$
|
show 2 more comments
$begingroup$
I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}
From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?
derivatives
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
$endgroup$
$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42
|
show 2 more comments
$begingroup$
I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}
From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?
derivatives
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
$endgroup$
I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}
From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?
derivatives
derivatives
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
edited Jan 18 at 15:24
Thomas Shelby
4,8082727
4,8082727
asked Jan 18 at 13:25
JakcjonesJakcjones
828
asked Jan 18 at 13:25
JakcjonesJakcjones
828
asked Jan 18 at 13:25
JakcjonesJakcjones
828
828
$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42
|
show 2 more comments
$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42
$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
add a comment |
$begingroup$
I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:
$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$
$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
add a comment |
$begingroup$
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
add a comment |
$begingroup$
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
edited Jan 18 at 13:58
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
answered Jan 18 at 13:52
Thomas ShelbyThomas Shelby
4,8082727
4,8082727
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
add a comment |
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
Can you explain how you come from the 3rd equality left side to the 4th?
$endgroup$
– Jakcjones
Jan 18 at 14:16
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
$begingroup$
$sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
$endgroup$
– Thomas Shelby
Jan 18 at 14:25
1
1
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
$begingroup$
Ok, got it. Wasn't visualizing it. Ty
$endgroup$
– Jakcjones
Jan 18 at 14:27
add a comment |
$begingroup$
I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:
$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$
$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:
$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$
$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:
$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$
$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
$endgroup$
I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:
$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$
$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
edited Jan 18 at 14:08
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
answered Jan 18 at 13:31
DonAntonioDonAntonio
180k1495233
180k1495233
add a comment |
add a comment |
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$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37
$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49
$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56
$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39
$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42