Find the derivative.
11) f(t)=(1)/(sqrt(8t-3))
12) g(t)= ((4t-3)^2)/ ((5t+1)^4)
Thanks.
To find the derivative of a function, you can use the rules of differentiation. Here's how you can find the derivatives of the given functions:
11) f(t) = 1/√(8t - 3)
To find the derivative of f(t), you can use the chain rule. The chain rule states that if you have a function g(h(t)), the derivative is given by g'(h(t)) * h'(t).
In this case, g(t) = 1/√t, and h(t) = 8t - 3. Let's find the derivatives of g(t) and h(t) first:
g'(t) = -1/2t^(-3/2) (using the power rule for derivatives)
h'(t) = 8 (the derivative of 8t is simply 8)
Now, we can use the chain rule to find f'(t):
f'(t) = g'(h(t)) * h'(t)
= (-1/2(8t - 3)^(-3/2)) * 8
Simplifying this expression gives:
f'(t) = -4/(8t - 3)^2
Therefore, the derivative of f(t) is f'(t) = -4/(8t - 3)^2.
12) g(t) = ((4t - 3)^2) / ((5t + 1)^4)
To find the derivative of g(t), we can use the quotient rule. The quotient rule states that if you have a function f(t)/g(t), the derivative is given by (f'(t)g(t) - f(t)g'(t)) / (g(t))^2.
In this case,
f(t) = (4t - 3)^2
g(t) = (5t + 1)^4
Let's find the derivatives of f(t) and g(t) first:
f'(t) = 2(4t - 3)(4) (using the chain rule for derivatives)
g'(t) = 4(5t + 1)^3(5) (using the chain rule for derivatives)
Now, we can use the quotient rule to find g'(t):
g'(t) = (f'(t)g(t) - f(t)g'(t)) / (g(t))^2
= (2(4t - 3)(4)(5t + 1)^4 - (4t - 3)^2(4(5t + 1)^3(5))) / ((5t + 1)^4)^2
Simplifying this expression gives:
g'(t) = (8(4t - 3)(5t + 1)^4 - 4(4t - 3)^2(5t + 1)^3(5)) / (5t + 1)^8
Therefore, the derivative of g(t) is g'(t) = (8(4t - 3)(5t + 1)^4 - 4(4t - 3)^2(5t + 1)^3(5)) / (5t + 1)^8.
Please note that it is always a good practice to simplify the derivatives as much as possible, but the above expressions are the simplified derivatives for the given functions.
These are standard formulas...
f()= 1/u
f'=-u'/u^2 where u is sqrt(8t-3), u'=4/sqrt( )
g= u/v
g'=-uv'/v^2 + u'/v where u=(4t-3)^2 v=(5t+1)^4
I assume you can do it from here.