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

william1941 | Student

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).

giorgiana1976 | Student

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)

neela | Student

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)