What is the derivative of the square root of (cos6x+sin(x^2)) step by step

We'll take the inside of the square root to be u and do this with the chain rule.

cos(6x) + sin(x^2) = u -> du/dx = 2xcos(x^2) – 6sin(6x)

1. (1/2)u^(-1/2) *du/dx

2. (1/2)(cos(6x) + sin(x^2))^(-1/2) (2xcos(x^2) – 6sin(6x))

3. (3sin(6x) – xcos(x^2))/sqrt(cos(6x) + sin(x^2))

There you have it, hope that helped.

cos(6x) + sin(x^2) = u -> du/dx = 2xcos(x^2) – 6sin(6x) is right as

d/dx* (cos nx) = -n sin nx so it is 6 sin 6x and not 3 sin 6x

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You need to use the chain rule in this case to solve for your derivative. This will probably a long derivative because there are functions within a function.

y = f(g(x))
y' = f'(g(x)) * g'(x)

In your case,

g(x) = cos(6x) + sin(x^2)
f(g(x)) = sqrt [ cos(6x) + sin(x^2) ]

y = f(g(x))
y = sqrt [ cos(6x) + sin(x^2) ]
y' = (1/2) [ cos(6x) + sin(x^2) ] ^ (-1/2) * [ cos(6x) + sin(x^2) ] ' ; Note: the derivative symbol still ( ' )
y' = (1/2) [ cos(6x) + sin(x^2) ] ^ (-1/2) * [ -6sin(6x) + (2x)(cos(x^2)) ]

After this, you can try simplifying it.

y' = (1/2) [ cos(6x) + sin(x^2) ] ^ (-1/2) * [ -6sin(6x) + (2x) ( cos(x^2) ) ]
y' = { [ (x) ( cos(x^2) ) - sin(6x) ] } / { sqrt [ cos(6x) + sin(x^2) ] }

use the chain rule. (1/2)[(cos6x +sin(x^2))^-1/2] times the derivitive of the inside
the derivitive of the inside => use the sum rule (the derivitive of a sum is the sum of the derivitives.)