# Find dy/dx when (A) y= sqrtlnx , (B) y= arctan 5x ?

Apr 5, 2018

(A) $\frac{d}{\mathrm{dx}} \sqrt{\ln} x = \frac{1}{2 x \sqrt{\ln x}}$

(B) $\frac{d}{\mathrm{dx}} \arctan 5 x = \frac{5}{25 {x}^{2} + 1}$

#### Explanation:

We seek the derivatives of:

(A) $y = \sqrt{\ln} x$
(B) $y = \arctan 5 x$

We require some standard derivatives:

 {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (af(x), af'(x), a " constant"), (x^n, nx^(n-1), n " constant (Power rule)"), (tan^(-1)x, 1/(1+x^2), ), (lnx, 1/x, ), (f(g(x)), f'(g(x)) \ g'(x),"(Chain rule)" ) :}

Part (A):

Applying the chain rule, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \sqrt{\ln} x$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{\frac{1}{2}}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(\ln x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{\ln x}} \frac{1}{x}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2 x \sqrt{\ln x}}$

Part (B):

Again, applying the chain rule, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \arctan 5 x$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{{\left(5 x\right)}^{2} + 1} \frac{d}{\mathrm{dx}} \left(5 x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{25 {x}^{2} + 1} 5$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{5}{25 {x}^{2} + 1}$