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Answer 1 · 2018-02-25T21:21:34

I will assume that the base of $log (x)$ $log(x)$ is $e$ $e$ and I will use $ln (x)$ $ln(x)$ in my answer.

$dy/dx = y'(x) = -(12*cos(ln((2x + 3)/(3- 2x))))/(4x^2 - 9)$

$f(x) = sin(log(x)) and y = f((2x +3) / (3 - 2x))$

I will assume that the base of $log(x)$ is $e$ and I will use $ln(x)$ in my answer.

$y$ is a composition of functions, so let's add some notation to make this more obvious.

Let:

$p(x) = sin(x)$
$q(x)= ln(x)$
$u(x) = 2x + 3$
$v(x) = 3- 2x$

then:

$y(x) = p(q((u(x))/(v(x))))$

We could attack this directly, but a property of logs can simplify it a bit:

$q((u(x))/(v(x))) = ln((u(x))/(v(x))) = ln(u(x)) - ln(v(x))$

$u'(x) = 2$
$v'(x) = -2$

$q'(x) = (u'(x))/(u(x)) - (v'(x))/(v(x)) = 2/(2x + 3) - (-2)/(3- 2x)$

$= 2/(2x + 3) - 2/(2x- 3) = (4x - 6 -(4x + 6))/(4x^2 - 9)$

$= -(12)/(4x^2 - 9)$

Applying the Chain Rule to y(x):

$y'(x) = p'(q(x)) * q'(x)$

$p'(x) = cos(x)$

so:

$y'(x) = cos(ln((2x + 3)/(3- 2x))) * (-(12)/(4x^2 - 9))$

$= -(12*cos(ln((2x + 3)/(3- 2x))))/(4x^2 - 9)$

Answer 2 · 2018-02-26T03:37:27

I have solved this way:
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