how do you find the derivative of
square root of 5 /t^5
f=a t^-n
f'= -an t^(-n-1)
i already know this rule but im still having trouble solving it
(5/(t^5))^1/2
you can rewrite this as:
(5t^(-5))^1/2
therefore,,
(1/2)*(-25t^-6)*(5t^(-5))^(-1/2)
so there,, =)
To find the derivative of the function f(t) = √(5 / t^5), you can use the Quotient Rule of differentiation. The Quotient Rule states that if you have a function f(x) = g(x) / h(x), where g(x) and h(x) are both differentiable functions, the derivative of f(x) is given by:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
In the case of f(t) = √(5 / t^5), we can let g(t) = √5 and h(t) = t^5. Let's differentiate these functions first:
g'(t) = 0 (since √5 is a constant)
h'(t) = 5t^4
Now, we can use the Quotient Rule to find f'(t):
f'(t) = (t^5 * 0 - √5 * 5t^4) / (t^5)^2
= (-5√5t^4) / (t^10)
= -5√5 / t^6
Therefore, the derivative of √(5 / t^5) is -5√5 / t^6.