Find the derivative: h(x)=7x/sqrt(5-2x)
Using the chain rule...I figured out my previous post.
To find the derivative of the function h(x) = 7x/√(5-2x), you can use the quotient rule combined with the chain rule.
The quotient rule states that if you have a function f(x) = g(x)/h(x), then its derivative f'(x) can be found using the formula:
f'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2.
In this case, g(x) = 7x and h(x) = √(5-2x).
Now, let's find the derivatives of g(x) and h(x).
The derivative of g(x) = 7x with respect to x is simply g'(x) = 7.
The derivative of h(x) = √(5-2x) requires the chain rule. Let u(x) = 5-2x. The derivative of u(x) with respect to x is u'(x) = -2.
Now, we can apply the chain rule to find the derivative of h(x):
h'(x) = [1/(2√u(x))] * u'(x) = (-2)/(2√(5-2x)) = -1/√(5-2x).
Finally, substitute these values into the quotient rule formula to find the derivative of h(x):
h'(x) = [(7)(√(5-2x)) - (7x)(-1/√(5-2x))]/[√(5-2x)]^2.
Simplifying this expression will give you the derivative of h(x) with respect to x.
h(x) = 7x *(5-2x)^-1/2
dh/dx = 7 (5-2x)^-1/2
+ (7x)d/dx(5-2x)^-1/2
Use the chain rule on the second term
d/dx(5-2x)^-1/2 = (-1/2)(-2)(5-2x)^-3/2
= (5-2x)^-3/2