Find the derivative. Simplify as much as possible.
s(x)=(9-12x)/[(x/3)-(1/4)]
Can someone explain this to me
let s= u/v
then s'= d(u/v)= u'/v -uv'/v^2
You can easily get u'and v'from the expression
u= 9-12x u'=-12
v= x/3-14 v'=1/3
check my thinking.
Actually the function reduces to
s(x) = -36, x not equal to 3/4
so s'(x) = 0
To find the derivative of the given function s(x)= (9-12x)/[(x/3)-(1/4)], we can use the quotient rule.
The quotient rule states that if we have a fraction of two functions f(x) and g(x), then the derivative is given by:
f'(x)g(x) - f(x)g'(x)
---------------------------
[g(x)]^2
In our case, f(x) = (9-12x) and g(x) = [(x/3)-(1/4)].
Now, let's find the derivatives of f(x) and g(x) separately.
The derivative of f(x) = (9-12x) can be found by applying the power rule and the constant rule. The power rule states that if we have a function u = ax^n, then the derivative of u with respect to x is given by du/dx = anx^(n-1), where a is a constant. In our case, when n = 1, the derivative becomes:
f'(x) = -12
Next, let's find the derivative of g(x) = [(x/3)-(1/4)].
The derivative of g(x) = (x/3)-(1/4) can be obtained by applying the power rule and the constant rule. Using the power rule, we have:
g'(x) = (1/3) - 0 = 1/3
Now, we can apply the quotient rule to find the derivative of s(x).
s'(x) = [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2
Plugging in the values we found earlier, we have:
s'(x) = [-12 * ((x/3)-(1/4))] / [((x/3)-(1/4))]^2
To simplify further, we can multiply numerator and denominator by the least common denominator, which is 12, to get rid of the fractions:
s'(x) = [-12 * (4x/12 - (3/12))] / [(4x/12 - (3/12))]^2
s'(x) = [(-48x + 36)] / [(4x - 3)]^2
And this is the simplified derivative of the given function s(x).