# `G(r) = sqrt(r) + root(3)(r)` Find the first and second derivatives of the function.

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### 3 Answers

Given G(r) = r ^1/2 + r ^1/3

Derive on both sides with respect to r

G'(r) = 1/2(r^-1/2) + 1/3 (r^-2/3)

Now derive with respect to r on both sides

G''(r) = -1/4(r^-3/2) + -2/9 (r^-5/3)

Thus we can get the first and second derivatives of r

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**Note:- **

**1) If f(x) = x^n ; where n = a real number**

**then f'(x) = n*x^(n-1)**

**2) If f(x) = k ; where k = real number ; then**

**f'(x) = 0**

Now, given G(r) = {r^(1/2)} + {r^(1/3)}

Thus, G'(r) = (1/2)*{r^(-1/2)} + (1/3)*{r^(-2/3)}

f"(x) = -(1/4)*{r^(-3/2)} + (-2/9)*{r^(-5/3)}

or, f"(x) = -[(1/4)*{r^(-3/2)} + (2/9)*{r^(-5/3)}]

`G(r)=r^(1/2)+r^(1/3)`

`G'(r)=(1/2)r^((1/2)-1)+(1/3)r^((1/3)-1)`

`=1/(2sqrt(r))+1/(3r^(2/3))`

`G'(r)=(1/2)r^(-1/2)+(1/3)r^(-2/3)`

`G''(r)=(1/2)(-1/2)r^((-1/2)-1)+(1/3)(-2/3)r^((-2/3)-1)`

`=-(1/4)r^(-3/2)-(2/9)r^(-5/3)`

`G''(r)=-(1/4)r^(-3/2)-(2/9)r^(-5/3)`

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