From the definition,
f¹ (x) = Lin f(x+h) - f(x)
Solution
f(x) =√sinx
f¹(x) = f(x+h) - f(x) \ h
√sin (x+h) - √sinx\h
let,
sin (x+h) = u+k
f sin x = u
k = (u+k) - u........i
= sin(x+h) - sin x........ii
h......0 = k......0...........iii
√u+k -√u\h
√u+k-√u\x. • k\h
√u+k-√u\k sin (x+h) - sinx\h
{(√u+k)²-(√u)²\k√u+k+√u}
2cos(x+h/2) sin (h/2)\h
1\√u+√u • 1 cos(x+0)
¹/2√u cos x
¹/2√sinx cos x (u= sin x)
f¹(x) = cos x/2√sinx
Therefore, the answer is cos x/2√sinx.
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