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

Answer 1

Emilia Aguilar

(A) # d/dx sqrtlnx = 1/(2xsqrt(lnx)) #
(B) # d/dx arctan 5x = 5/(25x^2+1)#

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)#

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Answer 2

William Abdalla

(A) ( \frac{dy}{dx} = \frac{1}{2x\sqrt{\ln{x}}} )

(B) ( \frac{dy}{dx} = \frac{5}{1+(5x)^2} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.