Question

Hey...SOLVE THIS.....​

Answer 1

d( √sinx)/ dx

First differentiate assuming sinx as x

So its 1/2√sinx

Now as its sin x So differentiate sinx now which is cosx

cosx /2√sinx

Now from first principle

y= √sinx

lim h->0. √sin( x+ h) -√ sinx)/ h

√sin ( x+ h)/h - √sinx/h

√( sinx cos h + cos x sinh)/ h -( √(sinx/h) × 1/√h

√( sinx cosh/h + ( cos x sinh )/h) × 1/√h

- √( sinx/h) × 1/√h

√( sinx/h) + cosx ) - √( sinx/h)))/√h

√( sin x/h + cos x) -√( sinx /h)) × √( sinx/h + cos x ) + √( sinx/h)) /√h( √sinx /h + cos x - √sinx /h))

= sinx/h + cos x - sinx /h )/ √( sinx/h + cos x) + √( sinx /h ) √h

= cos x)/ √h ( (√( sin x/h + cos x) + √( sin x/h))

= cos x)/√ ( sinx + cos x h ) + √sinx)

= cos x)/ √( sinx ) + √sinx

= cos x)/ 2√sinx

Answer 2

$\mathbb{\red{\huge{HI\:KAASHIKA}}}$

$\boxed{HERE\:IS\:YOUR\:ANSWER}$

$\mathfrak{\green{ANSWER:}}$

d( √sinx)/ dx

First differentiate assuming sinx as x

So its 1/2√sinx

Now as its sin x So differentiate sinx now which is cosx

cosx /2√sinx

Now from first principle

y= √sinx

lim h->0. √sin( x+ h) -√ sinx)/ h

√sin ( x+ h)/h - √sinx/h

√( sinx cos h + cos x sinh)/ h -( √(sinx/h) × 1/√h

√( sinx cosh/h + ( cos x sinh )/h) × 1/√h

- √( sinx/h) × 1/√h

√( sinx/h) + cosx ) - √( sinx/h)))/√h

√( sin x/h + cos x) -√( sinx /h)) × √( sinx/h + cos x ) + √( sinx/h)) /√h( √sinx /h + cos x - √sinx /h))

= sinx/h + cos x - sinx /h )/ √( sinx/h + cos x) + √( sinx /h ) √h

= cos x)/ √h ( (√( sin x/h + cos x) + √( sin x/h))

= cos x)/√ ( sinx + cos x h ) + √sinx)

= cos x)/ √( sinx ) + √sinx

= cos x)/ 2√sinx