# Hard Derivative.

• Oct 22nd 2009, 10:00 PM
mybrohshi5
Hard Derivative.
y=sqrt(10^(5-x))

find y'

how would you go about doing this one?

im studying for my "mastery test" on monday and i have never seen anything raise to a number - x

could someone show me how to do this please?

thank you
• Oct 22nd 2009, 10:02 PM
Chris L T521
Quote:

Originally Posted by mybrohshi5
y=sqrt(10^(5-x))

find y'

how would you go about doing this one?

im studying for my "mastery test" on monday and i have never seen anything raise to a number - x

could someone show me how to do this please?

thank you

Note that $\sqrt{10^{5-x}}=10^{\frac{1}{2}(5-x)}$. Then use the fact that $\frac{\,d}{\,dx}a^u=a^u\ln a\cdot\frac{\,du}{\,dx}$.

Can you take it from here?
• Oct 22nd 2009, 10:07 PM
mybrohshi5
yeah i can. but i seem to get a different answer then the answer of my review sheet.

i get -10^(.5(5-x)) * ln(10)

but....

the answer on the sheet is -10^(.5(5-x)) / 2
• Oct 22nd 2009, 10:08 PM
endling
Use logarithmic differentiation.
sqrt of 10^(5-x) is the same as 10^[(5-x)/2]
Take the natural log of both sides.
ln y = ln 10^[(5-x)/2]
ln y = [(5-x)/2] ln 10 because lnx^a = a lnx
then differentiate:
(1/y)(dy/dx) = (-1/2)ln10
solve for dy/dx:
dy/dx = (-1/2)(ln10)(10^[(5-x)/2])

edit: The answer sheet is wrong? :P
• Oct 22nd 2009, 10:18 PM
mybrohshi5
haha ok thank you very much =)