Question #765ea

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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.!