How do i find the derivative of
f(t)= 20- 30/rooot 9 + t^2
heres how i did it
f'(x)= 20' - 30/root(9+t^2) '
= 0 - 30/root(9+t^2) '
(Quotient rule)
f'(x)= 30' (root(9+t^2) - 30(root(9+t^2) '
f'(x)= -30(0.5(9+t^2)^-0.5) (2t)
the answer at the back of the book is 30t/(9+t^2)^1/2.... but i don't knohow to get it.
To find the derivative of the function f(t) = 20 - 30/√(9 + t^2), you are on the right track using the quotient rule. However, there seems to be a slight error in your calculation. Here's the correct step-by-step process:
Step 1: Apply the quotient rule.
The quotient rule states that if you have a function in the form h(x) = g(x)/[u(x)], where g(x) and u(x) are both differentiable functions, then the derivative of h(x) is given by:
h'(x) = [g'(x)*[u(x)] - g(x)*[u'(x)]]/[u(x)]^2
Let's apply this concept to your function f(t) = 20 - 30/√(9 + t^2).
g(t) = 20, u(t) = √(9 + t^2)
Step 2: Find the derivatives of g(t) and u(t).
g'(t) = 0 (since the derivative of a constant is zero)
u'(t) = 0.5(9 + t^2)^(-0.5) * 2t (using the chain rule)
Step 3: Plug the derivatives into the quotient rule formula.
f'(t) = [g'(t)*u(t) - g(t)*u'(t)]/[u(t)]^2
= [0 * √(9 + t^2) - 20 * (0.5(9 + t^2)^(-0.5) * 2t)]/[(√(9 + t^2))^2]
= [-20 * (0.5(9 + t^2)^(-0.5) * 2t)]/(9 + t^2)
Step 4: Simplify the expression.
f'(t) = [-20t * (0.5(9 + t^2)^(-0.5))]/(9 + t^2)
= [-20t/(9 + t^2)] * (0.5(9 + t^2)^(-0.5))
= [-20t/(9 + t^2)] * [1/(2√(9 + t^2))]
= [-20t/(2*(9 + t^2)√(9 + t^2))]
Simplifying further:
f'(t) = -10t/(√(9 + t^2)(9 + t^2))
So, the correct derivative of f(t) = 20 - 30/√(9 + t^2) is f'(t) = -10t/(√(9 + t^2)(9 + t^2)).