If f(t) = square root of 7/t^5. Find f'(x) and f'(2)
f= (7t^-5)^1/2
f'= 1/2 ( )^-1/2 * (-5*7t^-6)
the answer is not correct. I input
f'= 1/2 (7t^-5)^-1/2 * (-5*7t^-6)and its incorrect. Did I input the wrong answer?
To find the derivative of f(t), we can use the power rule and the chain rule.
To begin, let's rewrite the function as f(t) = 7 / (t^5)^(1/2). The derivative of this function can be found by applying the chain rule.
Step 1: Apply the power rule.
The derivative of t^n with respect to t is n*t^(n-1).
f'(t) = 7 * (1/2) * (t^(-5 - 1/2))
= 7/2 * t^(-12/2 - 1/2)
= 7/2 * t^(-13/2)
Step 2: Simplify the expression.
To simplify this expression, we can rewrite t^(-13/2) as 1 / t^(13/2).
f'(t) = 7/2 * 1 / t^(13/2)
= 7 / (2 * t^(13/2))
So, the derivative of f(t) is f'(t) = 7 / (2 * t^(13/2)).
Now, let's find f'(2) by substituting t = 2 into the derivative we found.
f'(2) = 7 / (2 * 2^(13/2))
= 7 / (2 * sqrt(2^13))
= 7 / (2 * sqrt(8192))
= 7 / (2 * 64)
= 7 / 128
= 0.0546875
Therefore, f'(2) is approximately 0.0546875.