1.

Using the definition of derivative find the derivative of `sqrt(sin x)`

Answer» Correct Answer - `(cos x)/(2sqrt(sin x))` (b) `-sin 2x` (c) `(1)/(sqrt(1-x^(2)))`
(a)`"Let "f(x)=sqrt(sin x)`
`therefore" "f(x+h)=sqrt(sin(x+h))`
`therefore" "(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(sqrt(sin(x+h))-sqrt(sin x))/(h)`
`=underset(hrarr0)lim(sin (x+h)-sin x)/(h(sqrt(sin(x+h)+sqrt(sinx))))`
`=underset(hrarr0)lim(2sin""((h)/(2))cos""((2x+h)/(2)))/(h(sqrt(sin(x+h)+sqrt(sin x))))`
`=underset(hrarr0)lim((sin""(h)/(2)))/(((h)/(2)))underset(hrarr0)lim(cos(x+(h)/(2)))/((sqrt(sin (x_+h)+sqrt(sin x))))`
`=(cos x)/(sqrt(sin x )+sqrt(sin x))=(cos x)/(2sqrt(sin x))`
(b) Let `f(x) =cos^(2)x.`
`therefore" "f(x+h)=cos^(2)(x+h)`
`therefore" "(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(cos^(2)(x+h)-cos^(2)x)/(h)`
`=underset(hrarr0)lim(sin^(2)x-sin^(2)(x+h))/(h)`
`=underset(hrarr0)lim(sin (x+x+h)sin (x-(x+h)))/(h)`
`=underset(hrarr0)lim(sin(2x+h)sin(-h))/(h)`
`=-underset(hrarr0)lim(sin h)/(h).underset(hrarr0)limsin(2x+h)`
`=-sin 2x`
(c) `(d(sin^(-1)x))/(dx)=underset(hrarr0)lim(sin^(-1)(x+h)-sin^(-1)x)/(h)`
`=underset(hrarr0)lim(sin^(-1)[(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])/(h)`
`=underset(hrarr0)lim(sin^(-1)[(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])/([(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])`
`xx((x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^2))/(h)`
`=1xxunderset(hrarr0)lim((x+h)^(2)(1-x^(2))-x^(2)(1-(x+h)^(2)))/(h[(x+h)sqrt(1-x^(2))+xsqrt(1-(x+h)^(2))])`
`=underset(hrarr0)lim((x+h)^(2)-x)/(h[(x+0)sqrt(1-x^(2))+xsqrt(1-(x+0)^(2))])`
`=underset(hrarr0)lim(h(2x+h))/(h[2xsqrt(1-x^(2))])`
`=underset(hrarr0)lim(2x+h)/(2xsqrt(1-x^(2)))`
`=(1)/(sqrt(1-x^(2)))`


Discussion

No Comment Found

Related InterviewSolutions