Precise definition of a derivative
$f(x) = x+sqrt{x}+4$
Obviously the derivative is $1 +frac{1}{2sqrt{x}}$
My problem is with using the precise definition of a limit namely:
$frac{d}{{dx}}fleft( x right) = mathop {lim }limits_{x to 0} frac{{fleft( {x + h} right) - fleft( x right)}}{h}$
How would you solve this using the precice definition of a derivative?
calculus
asked Nov 30 at 1:24
John
54
add a comment |
$f(x) = x+sqrt{x}+4$
Obviously the derivative is $1 +frac{1}{2sqrt{x}}$
My problem is with using the precise definition of a limit namely:
$frac{d}{{dx}}fleft( x right) = mathop {lim }limits_{x to 0} frac{{fleft( {x + h} right) - fleft( x right)}}{h}$
How would you solve this using the precice definition of a derivative?
calculus
asked Nov 30 at 1:24
John
54
1
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29
add a comment |
$f(x) = x+sqrt{x}+4$
Obviously the derivative is $1 +frac{1}{2sqrt{x}}$
My problem is with using the precise definition of a limit namely:
$frac{d}{{dx}}fleft( x right) = mathop {lim }limits_{x to 0} frac{{fleft( {x + h} right) - fleft( x right)}}{h}$
How would you solve this using the precice definition of a derivative?
calculus
asked Nov 30 at 1:24
John
54
$f(x) = x+sqrt{x}+4$
Obviously the derivative is $1 +frac{1}{2sqrt{x}}$
My problem is with using the precise definition of a limit namely:
$frac{d}{{dx}}fleft( x right) = mathop {lim }limits_{x to 0} frac{{fleft( {x + h} right) - fleft( x right)}}{h}$
How would you solve this using the precice definition of a derivative?
calculus
calculus
asked Nov 30 at 1:24
John
54
asked Nov 30 at 1:24
John
54
asked Nov 30 at 1:24
John
54
asked Nov 30 at 1:24
John
54
asked Nov 30 at 1:24
John
54
54
1
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29
add a comment |
1
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29
1
1
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29
add a comment |
1 Answer
1
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oldest
votes
begin{align*}frac{f(x+h)-f(x)}{h} &= frac{(x+h)+sqrt{x+h}+4-(x+sqrt{x}+4)}{h} \
&= frac{h + sqrt{x+h}-sqrt{x}}{h} \
&= 1+frac{sqrt{x+h}-sqrt{x}}{h} \
&= 1 + frac{(sqrt{x+h}-sqrt{x})(sqrt{x+h}+sqrt{x})}{h(sqrt{x+h}+sqrt{x})}\
&= 1 + frac{(x+h)-x}{h(sqrt{x+h}+sqrt{x})} \
&= 1 + frac{1}{sqrt{x+h}+sqrt{x}}
end{align*}
answered Nov 30 at 1:33
rogerl
17.4k22746
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
add a comment |
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1 Answer
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begin{align*}frac{f(x+h)-f(x)}{h} &= frac{(x+h)+sqrt{x+h}+4-(x+sqrt{x}+4)}{h} \
&= frac{h + sqrt{x+h}-sqrt{x}}{h} \
&= 1+frac{sqrt{x+h}-sqrt{x}}{h} \
&= 1 + frac{(sqrt{x+h}-sqrt{x})(sqrt{x+h}+sqrt{x})}{h(sqrt{x+h}+sqrt{x})}\
&= 1 + frac{(x+h)-x}{h(sqrt{x+h}+sqrt{x})} \
&= 1 + frac{1}{sqrt{x+h}+sqrt{x}}
end{align*}
answered Nov 30 at 1:33
rogerl
17.4k22746
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
add a comment |
begin{align*}frac{f(x+h)-f(x)}{h} &= frac{(x+h)+sqrt{x+h}+4-(x+sqrt{x}+4)}{h} \
&= frac{h + sqrt{x+h}-sqrt{x}}{h} \
&= 1+frac{sqrt{x+h}-sqrt{x}}{h} \
&= 1 + frac{(sqrt{x+h}-sqrt{x})(sqrt{x+h}+sqrt{x})}{h(sqrt{x+h}+sqrt{x})}\
&= 1 + frac{(x+h)-x}{h(sqrt{x+h}+sqrt{x})} \
&= 1 + frac{1}{sqrt{x+h}+sqrt{x}}
end{align*}
answered Nov 30 at 1:33
rogerl
17.4k22746
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
add a comment |
begin{align*}frac{f(x+h)-f(x)}{h} &= frac{(x+h)+sqrt{x+h}+4-(x+sqrt{x}+4)}{h} \
&= frac{h + sqrt{x+h}-sqrt{x}}{h} \
&= 1+frac{sqrt{x+h}-sqrt{x}}{h} \
&= 1 + frac{(sqrt{x+h}-sqrt{x})(sqrt{x+h}+sqrt{x})}{h(sqrt{x+h}+sqrt{x})}\
&= 1 + frac{(x+h)-x}{h(sqrt{x+h}+sqrt{x})} \
&= 1 + frac{1}{sqrt{x+h}+sqrt{x}}
end{align*}
answered Nov 30 at 1:33
rogerl
17.4k22746
begin{align*}frac{f(x+h)-f(x)}{h} &= frac{(x+h)+sqrt{x+h}+4-(x+sqrt{x}+4)}{h} \
&= frac{h + sqrt{x+h}-sqrt{x}}{h} \
&= 1+frac{sqrt{x+h}-sqrt{x}}{h} \
&= 1 + frac{(sqrt{x+h}-sqrt{x})(sqrt{x+h}+sqrt{x})}{h(sqrt{x+h}+sqrt{x})}\
&= 1 + frac{(x+h)-x}{h(sqrt{x+h}+sqrt{x})} \
&= 1 + frac{1}{sqrt{x+h}+sqrt{x}}
end{align*}
answered Nov 30 at 1:33
rogerl
17.4k22746
edited Nov 30 at 1:39
answered Nov 30 at 1:33
rogerl
17.4k22746
answered Nov 30 at 1:33
rogerl
17.4k22746
answered Nov 30 at 1:33
rogerl
17.4k22746
17.4k22746
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
add a comment |
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
2
2
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
I'd say the key step was when you multiplied the top and the bottom by $sqrt{x+h} + sqrt{x}$, which led to a nice simplification. It might not seem like an obvious thing to do, but, this is a standard trick that is sometimes taught in high school algebra courses.
– littleO
Nov 30 at 1:37
add a comment |
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1
have you tried evaluating the quantity? what difficulty did you face?
– Siong Thye Goh
Nov 30 at 1:29