Question

Need help in differentiation. Differentiate w.r.t. x. 2[sqrt(x)+2]^(1/2)

Answer 1

In an easier-to-understand form - \[2(\sqrt{x}+2)^.5\]

Answer 2

idk:(

Answer 3

f(x)=2[sqrt(x)+2]^(1/2)
f '(x)=2(1/2)(sqrtx+2)^-(1/2) D(sqrtx +2)
=[1/(sqrtx +2)^(1/2)] (1/2)x^-(1/2)
=[1/(sqrtx +2)^(1/2)] (1/2)x^-(1/2)
=1/[sqrtx (sqrtx +2)^.5]

Answer 4

\[\Large \begin{array}{l}
2\sqrt {\sqrt x + 2} \\
= 2\sqrt {{x^{1/2}} + 2} \\
= 2{\left( {{x^{1/2}} + 2} \right)^{1/2}}\\
f(x) = 2{(x)^{1/2}}\\
f'(x) = {x^{ - 1/2}}\\
g(x) = \sqrt x + 2 = {x^{1/2}} + 2\\
g'(x) = \frac{1}{2}{x^{ - 1/2}}\\
\frac{d}{{dx}} = {\left( {{x^{1/2}} + 2} \right)^{ - 1/2}} \cdot \frac{1}{2}{x^{ - 1/2}}\\
= \frac{1}{2} \cdot \frac{1}{{\sqrt {{x^{1/2}} + 2} }} \cdot \frac{1}{{\sqrt x }}\\
= \frac{1}{{2\sqrt {\sqrt x + 2} \cdot \sqrt x }}
\end{array}\]

Answer 5

f(x)=2[sqrt(x)+2]^(1/2)
f '(x)=2(1/2)(sqrtx+2)^-(1/2) D(sqrtx +2)
=[1/(sqrtx +2)^(1/2)] (1/2)x^-(1/2)
=[1/(sqrtx +2)^(1/2)] (1/2)x^-(1/2)
=1/[2sqrtx (sqrtx +2)^.5]