Question

Using the principal, find the derivative of
$\tan( \sqrt{x} )$
​

Answer 1

Answer:

Using the principal, find the derivative of $\tan( \sqrt{x} )$. Answer . vincy95.

Answer 2

Answer:

Let y=tan√x

Let δy be an increment in y, corresponding to an increment δx in x.

Then y+δy=tan(√x+δx)

=> Δy = tan(√x+Δx) - y

⇒δy=tan(√x+δx)−√tan√x

⇒δy=sin(√x+δ)xcos(√x+δx)−sin√xcos√x

⇒δy=sin(√x+δx)×cos√x−cos(√x+δx)×sinx√(cos√x+δx)(cos√x)

⇒δy=sin[√x+δx−√x][(co√sx+δx)](cos√x)

⇒δyδx=sin[√(x+δx)−√x]δx×1/(cos√x+δx)(cos√x)

⇒δy/δx=sin[√(x+δx)−√x]/x+δx−x ×1/[(cos√(x+δx)](cosx√)

⇒δy/δx=sin[√(x+δx)−√x](√x+δx)^2−(√x)^2×1/[cos(√x+δx)](cos√x)

⇒δy/δx=sin(√x+δx−√x)[√x+√(δx−x)]×1/(√x+√δx+x)×1/[cos(√x+δx)](cosx√)

⇒limδx→0

δy/δx= limδx→0 [sin(√x+δx−√x)/(x+δx√−x√)×1(x+δx√+x√)×1(cosx+δx√)(cos√x)]

⇒dydx=limδx→0 sin(x+δx√−x√)(√x+δx−√x)×limδx→0 1/(√x+δx+√x)×limδx→0 1/(cos√x+δx)(cosx√)

⇒dydx=1×1/0√x+0+√x ×1/cos√(x+0).cos√x

⇒dydx=12x√×1cos2x√

⇒dydx=sec^2√x/2√x