Using the chain rule, differentiate the function f(x)=square root(5+16x-(4x)squared).  what is the derivative of the square root?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = sqrt(5+16x - (4x)^2)

f(x) = sqrt(5+16x-16x^2)

Now let us factor:

f(x) = sqrt(-4x +5)(4x+1)

       = sqrt(-4x+5)*sqrt(4x+1)

Then, let f(x) = u*v

u= sqrt(-4x+5)  ==> u' = -2/sqrt(-4x+5)

v= sqrt(4x+1) ==>  v'= 2/sqrt(4x+1)

==> f'(x) = u'v + uv'

              = -2*sqrt(4x+1)/sqrt(-4x+5) + 2*sqrt(-4x+5)/sqrt(4x+1)

               = [-2(4x+1) + 2(-4x+5)]/sqrt(-4x+5)(4x+1)

                = (-8x -2 -8x +10)/sqrt(5+16x-4x^2)

                = (-16x +8)/sqrt(5+16-4x^2)

       f'(x)   = -8(x-2)/sqrt(5+16x-4x^2)

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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As per chain rule of differentiation:

dy/dx = (dy/du)(du/dv)(dv/dx)

Where:

y = f(u)

u = f(v) and

v = f(x)

To differentiate the given function using this rule we proceed as follows.

Given:

y = [5 + 16x - (4x)^2]^1/2

Let:

u = 5 + 16x - (4x)^2

v = 4x

Then:

y = f(u) = u^1/2 and

u = f(v) = 5 + 4v - v^2

dy/du = (1/2)u^(-1/2)

= (1/2)(5 + 4v - v^2)^(-1/2)

= (1/2)[5 + 16x - (4x)^2]^(-1/2)

= (1/2)[5 + 16x - 16x^2]^(-1/2)

du/dv = 4 - 2v

= 4 - 8x

dv/dx = 4

Then we calculate derivative of given expression as:

dy/dx = (dy/du)(du/dv)(dv/dx)

= {(1/2)[5 + 16x - (4x)^2]^(-1/2)}(4 - 8x)4

= 16(1 - 2x){(1/2)[5 + 16x - (4x)^2]^(-1/2)}

= 16(1 - 2x){(1/2)[5 + 16x - (4x)^2]^(-1/2)}

= 8(1 - 2x)(5 + 16x - 16x^2)^(-1/2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To differentiate the function, we'll have to calculate the derivative of the expression under the square root and, after that, we'll differentiate the square root.

 We'll note sqrt(5+16x-4x^2) = u(v(x))

Where u(v) = sqrt v

v(x) = (5+16x-4x^2)

Now, we'll differentiate v(x), with respect to x:

v'(x) = (5+16x-4x^2)'

v'(x) = (5)' + (16x)' - (4x^2)'

v'(x) = 0 + 16 - 8x

v'(x) = 16-8x

u(v(x))' = (sqrt v)' = v'(x)/2sqrt v

But u(v(x)) = f(x) =>

f'(x) = (16-8x)/2sqrt(5+16x-4x^2)

f'(x) = 8(2-x)/2sqrt(5+16x-4x^2)

f'(x) = (2-x)/sqrt(5+16x-4x^2)

f'(x) = (2-x)sqrt(5+16x-4x^2)/(5+16x-4x^2)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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To find the drivative of y = sqrt(5+16x-4x^2)

d/dxf(u) = {d/duf(u) }{du/dx}.

dy/dx  =  d/du {u^(1/2} {du/dx} , where u = 5+16x-x^2.

dy/dx = {(1/2)(5+16x-5x^2) ^(1/2-1) }d/dx{5+16x-x^2}

dy/dx = (1/2){5+16x-x^2}^(-1/2)*{16-2x}

dy/dx = (16-2x)/{2sqrt(5+16x-x^2)}

 

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