# Differentiate the function f(x) = ln sqrt ( 17x^2 - cosx), using the chain rule.

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### 3 Answers

Let A = f(x) = ln sqrt ( 17x^2 - cos x)

Let 17x^2 - cos x = y

So A = ln sqrt y

Let sqrt y = z

=> A = ln z

We know that : dA / dx = ( dA / dz ) * (dz / dy ) * (dy / dx)

Now ( dA / dz ) = diffrential of ln z = 1 / z

dz / dy = 1/ 2*sqrt y

dy / dx = 34 x + sin x

Therefore dA / dx = (1 / z) * ( 1/ 2*sqrt y) *( 34 x + sin x)

=> (1 / sqrt y) * ( 1/ 2*sqrt y) *( 34 x + sin x)

=> [1 / sqrt (17x^2 - cos x)] * [ 1/ 2*sqrt (17x^2 - cos x)] *( 34 x + sin x)

=> [ 1/ 2 (17x^2 - cos x) ] ( 34 x + sin x)

**The result is ( 34 x + sin x)/ 2 (17x^2 - cos x).**

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

We'll note ln sqrt ( 17x^2 - cosx)= u(v(t(x)))

Where u(v) = ln v

v(t) = sqrt t

t(x) = 17x^2 - cosx

Now, we'll differentiate t(x), with respect to x:

t'(x) = (17x^2 - cosx)'

t'(x) = (17x^2)' - (cosx)'

t'(x) = 34x - (-sin x)

t'(x) = 34x + sin x

v(t(x))' = (sqrt t)' = t'(x)/2sqrt t

v(t(x))' = (34x + sin x)/2sqrt (17x^2 - cosx)

u'(v) = (ln v)'

u'(v(t(x))) = v(t(x))'/v

u'(v(t(x))) = (34x + sin x)/2sqrt (17x^2 - cosx)*sqrt ( 17x^2 - cosx)

u'(v(t(x))) = (34x + sin x)/2(17x^2 - cosx)

But f'(x) = u'(v(t(x)))

**f'(x) = (34x + sin x)/2(17x^2 - cosx)**

To differentiate the function f(x) = lnsqrt(17x^2-cosx).

Solution:

d/dx u(v(x)) = (du/dv )(dv/dx) = is the chain rule of differentiation.

f'(x) = ln {sqrt(17x^2-cosx} = (1/u )* du/dx , where u = sqrt (17x^2-cosx).

f'(x) = {1/sqrt(17x^2-cosx)} {sqrt(17x^2-cosx}'

f'(x) = {1/sqrt(17x^2-cosx)} (1/2){17x^2-cosx}^(1/2-1) * {17x^2-cosx}

f'(x) = (1/2){(1/sqrt(17x^2-cosx)}{1/(17x^2-cosx)} (17x^2-cosx)'

f'(x) = (1/2) {1/(17x^2-cosx)^(3/2)}(2*17x+sinx)

f'(x) = (1/2)(34x+sinx)/(17x^2-cosx)^(3/2)