Question #765ea

Question

941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T09:55:07

#"The Reqd. Derivative="(sqrte)^sqrtx/(4sqrtx).#

Let #y=(sqrte)^(sqrtx)=(e^(1/2))^sqrtx=e^(1/2sqrtx)=e^t,# say, where, #t=1/2sqrtx#.

Thus, #y=e^t, t=1/2sqrtx=1/2x^(1/2),# whivh means that,

#y" is a function of "t, and, t" of "x.#

In such cases, we use the Chain Rule to find #dy/dx," i.e. to say,"#

#dy/dx=dy/dt dt/dx.....................(star)#

Now, #because, y=e^t, dy/dt=e^t,#

# and, t=1/2x^(1/2), dt/dx=1/2{1/2x^(1/2-1)}=1/4x^(-1/2)=1/(4sqrtx).#

Utilising all these in #(star), dy/dx=e^t/(4sqrtx)#

Reverting back from #t" to "x#, we finally get,

#"The Reqd. Derivative="e^(1/2sqrtx)/(4sqrtx)=(sqrte)^sqrtx/(4sqrtx).#

Enjoy Maths.!