Precise definition of a derivative












0














$f(x) = x+sqrt{x}+4$



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







share|cite|improve this question


















  • 1




    have you tried evaluating the quantity? what difficulty did you face?
    – Siong Thye Goh
    Nov 30 at 1:29


















0














$f(x) = x+sqrt{x}+4$



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







share|cite|improve this question


















  • 1




    have you tried evaluating the quantity? what difficulty did you face?
    – Siong Thye Goh
    Nov 30 at 1:29
















0












0








0







$f(x) = x+sqrt{x}+4$



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







share|cite|improve this question













$f(x) = x+sqrt{x}+4$



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






active

oldest

votes


















2














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*}






share|cite|improve this answer



















  • 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













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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*}






share|cite|improve this answer



















  • 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














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*}






share|cite|improve this answer



















  • 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








2






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*}






share|cite|improve this answer














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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 at 1:39

























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
















  • 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




















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