anonymous
  • anonymous
find the derivative of the function h(x)=sqrt(2-x)
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
find the derivative of the function h(x)=sqrt(2-x)
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
use chain rule
myininaya
  • myininaya
by the sqrt( ) can be written as ( )^(1/2)
anonymous
  • anonymous
we are just in the first section and haven't learned the chain rule, and are required to use a longer method

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
ok the definition then
anonymous
  • anonymous
right
myininaya
  • myininaya
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =f'(x)\]
myininaya
  • myininaya
where we have f(x)=sqrt(2-x) and so f(x+h)=sqrt(2-(x+h)) <--here just replaced my x's with (x+h)'s
myininaya
  • myininaya
now plug in
myininaya
  • myininaya
sometimes a good method when you are trying to evaluate a limit algebraically and you have a difference/sum of radicals rationalizing might help
myininaya
  • myininaya
to rationalize recall conjugates
myininaya
  • myininaya
for example the conjugate of a-b is a+b and the conjugate of a+b is a-b
myininaya
  • myininaya
do you want to try to see what happens if you do the following: \[\lim_{h \rightarrow 0} \frac{\sqrt{2-(x+h)}-\sqrt{2-x}}{h} \cdot \frac{\sqrt{2-(x+h)}+\sqrt{2-x}}{\sqrt{2-(x+h)}+\sqrt{2-x}}\]
anonymous
  • anonymous
right, I got it down to -1/sqrt2-x-h +sqrt2-x
myininaya
  • myininaya
one sec let me check
myininaya
  • myininaya
that looks pretty sweet
myininaya
  • myininaya
\[\lim_{h \rightarrow 0}\frac{-1}{\sqrt{2-x-h}+\sqrt{2-x}}\]
myininaya
  • myininaya
your job is to just replace h with 0
myininaya
  • myininaya
any you could do a little magic afterwards to make the answer look a bit more tidy
anonymous
  • anonymous
so -1/2sqrt2-x?
anonymous
  • anonymous
thank you!
myininaya
  • myininaya
\[\frac{-1}{2 \sqrt{2-x}} \text{ is \right!} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.