Question: Find the derivative of the function. f(t) = 7(+6 - 2) 5 Step 1 Recall that the derivative of a constant times a differentiable function

 Find the derivative of the function. f(t) = 7(+6 - 2)5 Step 1 Recall that the derivative of a constant times adifferentiable function is equal to the constant times the derivative of the

function. In other words, we have the following where c is aconstant. d [CA(X) ] = C- F(x ) ] Applying this rulewhen finding the derivative of the given function gives us the following

Find the derivative of the function. f(t) = 7(+6 - 2) 5 Step 1 Recall that the derivative of a constant times a differentiable function is equal to the constant times the derivative of the function. In other words, we have the following where c is a constant. d [CA(X) ] = C- F(x ) ] Applying this rule when finding the derivative of the given function gives us the following result. f' ( t ) = - [7 (+6 - 2) 5 ] = 79 at (+6 - 2) 5 (16 - 2) Step 2 Recall the general power rule, which states that if the function f is differentiable and h(x) = [f(x)]", where n is a real number, then the following is true. h' ( x ) = - [f (x) ]" = n[f ( x ) in - 1f ' ( x ) For -[(to - 2)"] we have a differentiable function being raised to a power. Let g(t) = to - 2 and n = 5. Therefore, we have the following. [(g(t))"] = - [(t6 - 2) 5 ] Applying the general power rule to the above gives us the following result. at [(9(t))"] = n[g(t)in - 1g'(t) at [(to - 2)5] = 5(+6 - 2)5- 1 +5 )+ 5 ( + 6 - 2 ) 4Enclosing the Largest Area The owner of the Rancho Grande has 2,980 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions (in yd) of the largest area he can enclose? shorter side yd longer side yd What is this area (in ydz)? yd2 Volume of a Cube The volume Vof a cube with sides of length X in. IS changing with respect to time. At a certain instant of time, the sides of the cube are 3 in. long and increasing at the rate of 0.5 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time? cu in/s

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