00:01
In this problem we are provided with the function f of t which equals to t squared plus 2 times t plus 8 the whole multiplied with 2 times t squared plus 6.
00:16
We are asked to find out f prime of t that is the first derivative of the function and we are also asked to find out f prime of 5.
00:26
That is we substitute the value of t as 5.
00:29
So first let us differentiate the given function.
00:32
We have f prime of t to be equal to.
00:35
Here we have the product of two functions and the product of two functions can be differentiated using the product rule that is uv prime plus v u prime.
00:47
So in this case we have t squared plus 2 times t plus 8 remains as it is differentiating the second term by making use of the derivative of t raised to the power n, which is n times t raised to the part n minus 1.
01:03
So we get 2 times 2 times t and the derivative of any constant is 0 plus now this time 2 times t squared plus 6 remains as it is and we differentiate the first term using the same formula so we get 2 times t plus 2 plus 0.
01:22
Simplifying this we have 4 times t times t squared plus 2 t plus 8 plus 2 t squared plus 6 times 2 times t plus 2.
01:37
So now let us further simplify this by multiplying the terms we get 4 times t cubed plus 8 times t squared plus 32 times t...