# Calculate the derivative of the function y = cos(x^4 + 2x^2)*ln[(x+7)/(x-7)]

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y= cos(x^4 + 2x^2) *ln (x+7)/(x-7)

let y = u*v such that:

u = cos(x^4 + 2x^2) ==> u' = -(4x^3 + 4x)*sin(x^4+ 2x^2)

v = ln (x+7) - ln (x-7) ==> v' = 1/(x+7) - 1/(x-7) = -14/(x^2 - 49)

==> y'= u'v + uv'

** y' = -(4x^3+4x)*sin(x^4+2x^2)*ln (x+7)/(x-7) + cos(x^4+2x^2* -14/(x^2-49)**

** **

We'll note f(x) = y.

We'll calculate the derivative of the composed function f(x) with respect to the fact that it contains a product, the function cosine and the logarithm of a quotient:

(u*v)' = u'*v + u*v'

f(x)' = {cos (x^4 + 2x^2)*ln[(x+7)/(x-7)]}'

f(x)' = [cos (x^4 + 2x^2)]'*ln[(x+7)/(x-7)] +cos (x^4 + 2x^2)*{ln[(x+7)/(x-7)]}'

We'll differentiate the logarithm:

{ln[(x+7)/(x-7)]}' = [(x-7)/(x+7)]*[(x-7-x-7)/(x-7)^2]

We'll eliminate and reduce like terms:

{ln[(x+7)/(x-7)]}' = -14/(x^2-49)

We'll differentiate [cos (x^4 + 2x^2)]':

[cos (x^4 + 2x^2)]' = -(x^4 + 2x^2)'*sin(x^4 + 2x^2)

[cos (x^4 + 2x^2)]' = -(4x^3+2x)*sin(x^4 + 2x^2)

**f'(x) = -(4x^3+2x)*sin(x^4 + 2x^2)*ln[(x+7)/(x-7)] - 14cos (x^4 + 2x^2)/(x^2 - 49)**

To calculate the derivative of:

y = cos(x^4 + 2x^2)*ln[(x+7)/(x-7)]

Right side is aproduct . So we use (u*v)' = u'*v+u*v'

And we use{ f(g(x)) }'= f'(g(x)*g'(x).

Where u = cos(x^2+2x^2), u' = -{sin(x^4+2x^2)}{x^4+2x^2}' = -{sin(x^4+2x^2)}(4x^3+4x)

v = ln(x+7)/(x-7), v'= {x-7)/(x+7){1(x-7)-(x+7)]/(x-7)^2}={ (x-7)/(x-7)}(-14/(x-7)^2}

Therefore

y' = -sin(x^4+2x^2)}{4x^3+4x)ln(x+7)/(x-7) - {cos(x^4+2x^2)}{(x-7)(x+7)(-14)/(x-7)^2}

y' = -4(x^3+x)sin (x^4+2x^2)} + {14/(x^2-49)}cos(x^4+2x^2)