# Take derivative d/dx (3tan^(-1)square root(x)) show steps

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Use the chain rule to differentiate this function.

d/dx (arctan(u)) = u'/(1 + u^2)

let u = x^(1/2) --> u' = (1/2)*x^(-1/2)

So,

d/dx (3arctan(u)) = 3u'/(1 + u^2) = (3/2)*x^(-1/2)/(1 + (x^(1/2))^2)

**= 3/( x^(1/2)*(1 + x) )**

To find the derivative of (3tan^(-1)square root(x)).

We know that d/dx(f(U(x))) ={ d/dUf(U)} = d/dx f(U)]*d/dx(U(x)).....(1)

(d/dx)(x^n) = nx^(n-1)........................................(2)

(d/dx)tan inverse x = 1/(1+x^2).............................(3)

d/dx[k*f(x)] = k* d/dx (f(x) or kf'(x)......................(4), where k is a constant.

We use the above in finding the derivative of the given expression.

Let y = (3tan^(-1)square root(x)) or

y=3 f(u(x)) .........................................................(5), where f(u) tan inverse u and u(x) = sqrtx = x^(1/2).

Therefore,

dy/dx=d/dx{3f(u(x)} = 3*d/dx{f(U) * d/dx(u(x)

=3 {1/(1+U^2)} ((1/2) x^(1/2 -1))

=3/(1+(sqrtx)^2) }(1/2)x^(-1/2)

=3/{2(1+x)x^(1/2)}