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3 votes
Use the definition:

f'(x)=
\lim_(h \to 0) (f(x+h)-f(x))/(h)
to find f'(x) for:
f(x)=
(1)/(√(x))+x
I need the WORK, not the answer. Thanks!

User Dumoko
by
6.4k points

1 Answer

2 votes

Using the given definition, for
f(x)=\frac1{\sqrt x}+x, we have


f'(x)=\displaystyle\lim_(h\to0)\frac{\left(\frac1{√(x+h)}+x+h\right)-\left(\frac1{\sqrt x}+x\right)}h

Right away, we see x and -x in the numerator, so we can drop those terms.


f'(x)=\displaystyle\lim_(h\to0)\frac{\frac1{√(x+h)}+h-\frac1{\sqrt x}}h

Remember that limits distribute over sums, i.e.


\displaystyle\lim_(x\to c)(f(x)+g(x))=\lim_(x\to c)f(x)+\lim_(x\to c)g(x)

so we can separate the h from everything else in the numerator:


f'(x)=\displaystyle\lim_(h\to0)\frac{\frac1{√(x+h)}-\frac1{\sqrt x}}h+\lim_(h\to0)\frac hh

Since h ≠ 0, we have
\frac hh=1, so the second limit is simply 1.


f'(x)=\displaystyle\lim_(h\to0)\frac{\frac1{√(x+h)}-\frac1{\sqrt x}}h+1

For the remaining limit, focus on the numerator for now. Combine the fractions in the numerator:


\frac1{√(x+h)}-\frac1{\sqrt x}=(\sqrt x-√(x+h))/(\sqrt x√(x+h))

Recall the difference of squares identity,


a^2-b^2=(a-b)(a+b)

Let
a=\sqrt x and
b=√(x+h). Multiply the numerator and denominator by
(a+b), so that the numerator can be condensed using the identity above.


(\sqrt x-√(x+h))/(\sqrt x√(x+h))\cdot(\sqrt x+√(x+h))/(\sqrt x+√(x+h))


=((\sqrt x)^2-(√(x+h))^2)/(\sqrt x√(x+h)(\sqrt x+√(x+h)))


=(x-(x+h))/(\sqrt x√(x+h)(\sqrt x+√(x+h)))


=-\frac h{\sqrt x√(x+h)(\sqrt x+√(x+h))}

Back to the limit: all this rewriting tells us that


f'(x)=\displaystyle\lim_(h\to0)\frac{-\frac h{\sqrt x√(x+h)(\sqrt x+√(x+h))}}h+1

Again, the h's cancel, and we can pull out the factor of -1 from the numerator and simplify the fraction:


f'(x)=\displaystyle-\lim_(h\to0)\frac1{\sqrt x√(x+h)(\sqrt x+√(x+h))}+1

The remaining expression is continuous at h = 0, so we can evaluate the limit by substituting directly:


f'(x)=-\frac1{\sqrt x√(x+0)(\sqrt x+√(x+0))}+1


f'(x)=-\frac1{2x\sqrt x}+1

or, if we write
\sqrt x=x^(1/2), we get


f'(x)=-\frac12x^(-3/2)+1

User Rady
by
6.3k points
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