Question

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!

Answer 1

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`