# Differentiate the function y=sin^4(sqrt(u))

kjcdb8er | Certified Educator

Here, you have one function inside another (inside another) ; so you need to use the chain rule (twice).

Let x = sqrt(u), f(x) = sin^4(x)

dy/du = df/dx * dx/du

df/dx = 4sin^3(x) * cos(x)   --> here, you used the chain rule to compute d/dx ( g^4), where g = sin x, and dg/dx = cos x.

dx/du = d/du ( u^(1/2)) = 1/2 u^(-1/2)

So,

dy/du = 4sin^3(x) * cos(x) * 1/2 u^(-1/2)

= 4sin^3(sqrt(u)) * cos(sqrt(u)) * 1/2 u^(-1/2)

hala718 | Certified Educator

y= sin^4(sqrtu)

Let t= sqrtu

==> dt= 1/2sqrtu du

==> y= sin^4 ( t)

==> dy = 4sin^3 (t) ( sint)'  dt

==> dy= 4sin^3 ( t) (cost) dt

==> dy = 4*sin^3 (t) * cos(t) dt

Now substitute with:

t= sqrtu

dt = 1/2sqrtu du

==> dy = 4 *sin^3(sqrtu) * cos(sqrtu) / 2sqrtu) du

==> dy/du = 2sin^3(sqrtu)*cos(sqrtu)/ sqrtu

neela | Student

To differentiate y = sin^4 (sqrt(u)).

This is like differentiating a function of a function.

If y= f(g(x)), then dy/dx = y' =  {f(g(x))}' = df(g(x)x)/dg(x)*dg(x)/dx = {df(g(x))/d(g(x))} g'(x).

Also we use d/dx(x^n) = nx^(n-1) to differentiate the given expression.

y = sin ^4(sqrtu).

Therefore y ' = {sin^4(sqrtx)} = 4sin ^3 (sqrt(x))* (sin(sqrtx))'

y' = 4sin^3(sqrtu)*cos(sqrtu)* (sqrtu)'

y' = 4sin^3 (sqrtu) * cos(sqrtu)* (1/2)(u^(-1/2).

y' = 2 [sin (sqrtu)]^3 [cos(sqrtu)]/x^(1/2).

troyhigh | Student

y = sin^4(sqrt(u))

y' = 4cos^3(sqrt(u))*-sin(sqrt(u))*1/(2*sqrt(u))

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