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

3 Answers | Add Yours

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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).

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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)

 

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question