Question

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

Answer 1

(A) `d/dx sqrtlnx = 1/(2xsqrt(lnx))`

(B) `d/dx arctan 5x = 5/(25x^2+1)`

Explanation:

We seek the derivatives of:

(A) `y= sqrtlnx`
(B) `y= arctan 5x`

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:

`dy/dx = d/dx sqrtlnx`

`\ \ \ \ \ \ = d/dx (lnx)^(1/2)`

`\ \ \ \ \ \ = 1/2(lnx)^(-1/2) d/dx (lnx)`

`\ \ \ \ \ \ = 1/(2sqrt(lnx)) 1/x`

`\ \ \ \ \ \ = 1/(2xsqrt(lnx))`

Part (B):

Again, applying the chain rule, we have:

`dy/dx = d/dx arctan 5x`

`\ \ \ \ \ \ = 1/((5x)^2+1) d/dx (5x)`

`\ \ \ \ \ \ = 1/(25x^2+1) 5`

`\ \ \ \ \ \ = 5/(25x^2+1)`