How do you find f'(x) using the definition of a derivative f(x)=6 x + 2 /sqrt{x}?
1 Answer
Determination, attention to detail and algebra. See below.
Explanation:
= lim_(hrarr0) ([6(x+h)+2/sqrt(x+h)] - [ 6x+2/sqrtx])/h
This limit looks hard to evaluate, but we can split it into two limits:
= lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h
The first limit evaluates to
This is a combination of problems you've probably seen before.
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We should try the methods that worked for those problems.
First, let's get a single rational expression in the numerator.
And now a single rational expression.
= lim_(hrarr0)((2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx))
If we try to evaluate the limit, we still get the indeterminate form
I suppose we could try rationalizing the numerator to see if that helps.
= lim_(hrarr0)((2sqrtx-2sqrt(x+h)))/(hsqrt(x+h)sqrtx) * ((2sqrtx+2sqrt(x+h)))/((2sqrtx+2sqrt(x+h)))
= lim_(hrarr0)((4x-4(x+h)))/(hsqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))
= lim_(hrarr0)(-4cancel(h))/(cancel(h)sqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))
The numerator no longer goes to
That means we can evaluate the limit. (Then we'll add the
= (-4)/ (sqrt(x+0)sqrtx (2sqrtx+2sqrt(x+0))
= (-4)/ (sqrtxsqrtx (4sqrtx)) = (-1)/(sqrtx)^3
Finally we finish
= lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h
= 6 - 1/(sqrtx)^3