Qn on Differentiation

wishboy SG

*10. Given that f(x) = sqrt[1 + sqrt(x)], where x is more than or equals to 0, show that f'(x) = 1/{4 sqrt[x + x sqrt(x)]}.

pls help me on this qn Sad

eagle SG

My strategy is I always convert to nice nice looking form then start doing

f(x) = (1 + x^0.5) ^ 0.5

so f'(x) = 0.5 * (1 + x^0.5)^(-0.5) * (d/dx {x^0.5} ) => chain rule I think
f'(x) = 0.5 * (1 + x^0.5)^(-0.5) * 0.5 * (x^-0.5)
f'(x) = 0.25 (x + x sqrt(x) ) ^ -0.5
f'(x) = 1/{4 sqrt[x + x sqrt(x)]} (shown)

wishboy SG

erm wad does f’(x) mean ar?

eagle SG

means differentiate f(x), or rather, a more accurate description is the first differential of f(x)

wishboy SG

oh ok
y must they write until like so chim lol
can jus write differentiate the following w.r.t. x mah Laughing

eagle SG

because it is a short form

and when you reach till higher differentials, it is an easier and shorter way of representing them

f''(x) is differentiate f(x) twice, etc

or the '' could be replaced by (3) or other numbers, which means 3rd differential or etc...

or

when you have many different types of function
f(x), g(x), h(x), etc

it is clearer and easier if you see differentials as
f'(x), g'(x), h'(x), etc...

wishboy SG

o...
i read ur post i alr confuse liao
so many functions

thanks for ur help !

eagle SG

erm... not impt de... just extra information...

For now, just take it as f'(x) is d/dx of f(x)

Slowly you will understand Very Happy