- WE'RE GIVEN H OF T = THE 
INTEGRAL OF SINE X TO THE 5TH
DIVIDED BY X TO THE 4TH 
FROM 2 TO THE SQUARE ROOT OF T.
WE WANT TO FIND H PRIME OF T 
OR THE DERIVATIVE OF H OF T.
TO DO THIS, WE'LL BE APPLYING
THE SECOND FUNDAMENTAL THEOREM 
OF CALCULUS STATED HERE BELOW
WHERE THE DERIVATIVE 
WITH RESPECT TO X
OF THE INTEGRAL OF F OF T FROM 
A TO X IS EQUAL TO F OF X.
WE CAN THINK OF THE DERIVATIVE 
AND INTEGRAL UNDOING EACH OTHER.
AND THEREFORE, TO EVALUATE THIS 
WE SUBSTITUTE THE VARIABLE X
INTO THE INTEGRANT FUNCTION 
FOR T GIVING US F OF X.
NOTICE HOW THE LOWER LIMIT 
OF INTEGRATION IS A CONSTANT
AND THE UPPER LIMIT INTEGRATION 
IS JUST X.
NOTICE IN OUR CASE THOUGH, 
WE HAVE THE SQUARE ROOT OF T.
BECAUSE THIS NOT JUST T,
WE'LL HAVE TO APPLY THE CHAIN 
RULE TO FIND H PRIME OF T.
SO LET'S BEGIN BY LETTING U 
= THE SQUARE ROOT OF T.
WELL, IF U IS EQUAL 
TO THE SQUARE ROOT OF T
WE COULD WRITE THIS AS 
TO THE 1/2.
AND THEREFORE, U PRIME, 
OR IF WE WANT UDT,
WOULD BE EQUAL TO 1/2 
x T TO THE -1/2,
WHICH WOULD BE THE SAME AS 1 
DIVIDED BY 2T TO THE +1/2.
NOW LET'S WRITE THIS 
IN TERMS OF U.
WE CAN NOW SAY THAT H PRIME OF U
WOULD BE EQUAL TO THE 
DERIVATIVE, WITH RESPECT TO U,
OF THE INTEGRAL FROM 2 TO U 
OF SINE X TO THE 5TH
DIVIDED BY X TO THE 4TH DX.
NOTICE IN THIS FORM 
WE CAN APPLY THE THEOREM
EXACTLY AS STATED HERE.
MEAN TO EVALUATE THIS 
WE CAN SUBSTITUTE U FOR X.
SO THIS WOULD GIVE US SINE U 
TO THE 5TH DIVIDED BY U TO 4TH.
BUT NOTICE HOW THIS IS 
H PRIME OF U
AND WE'RE LOOKING FOR 
H PRIME OF T.
THIS IS WHERE WE APPLY 
THE CHAIN RULE.
H PRIME OF T,
OR IF WE WANT DHDT, 
IS EQUAL TO DHDU OR H PRIME OF U
x DUDT OR U PRIME,
WHICH WE HAVE OVER HERE 
ON THE RIGHT.
SO THIS WOULD GIVE US 
SINE U TO THE 5TH
DIVIDED BY U TO THE 4TH x DUDT,
WHICH IS 1 DIVIDED BY 
2T TO THE 1/2.
BUT NOTICE HERE WE HAVE BOTH U's 
AND T's HERE,
WE WANT A DERIVATIVE FUNCTION 
IN TERMS OF ONLY T.
SO NOW WE'LL SUBSTITUTE 
T TO THE 1/2
OR THE SQUARE ROOT OF T FOR U.
SO THIS WOULD GIVE US SINE OF T 
THE 1/2 TO THE 5TH
DIVIDED BY T THE 1/2 TO THE 4TH 
x 2T TO THE 1/2.
SO LET'S CLEAN THIS UP 
ONE MORE TIME.
WE HAVE H PRIME OF T = SINE OF-- 
THIS WOULD BE T TO THE 5/2.
LET'S WRITE THIS 
AS A SQUARE ROOT.
SINCE THE DENOMINATOR 
WOULD BE 2,
IT WOULD BE THE SINE OF THE 
SQUARE ROOT OF T TO THE 5TH
DIVIDED BY--
HERE WE'D HAVE 2T SQUARED 
x T TO THE 1/2.
THAT'S T TO THE 2 1/2 
OR T TO THE 5/2 POWER,
WHICH, AGAIN, WOULD BE THE 
SQUARE ROOT OF T TO THE 5TH.
WE CAN PROBABLY LEAVE IT 
IN THIS FORM HERE,
BUT NOTICE HOW THE SQUARE ROOTS 
WOULD SIMPLIFY.
WE COULD WRITE THIS AS SINE OF--
THE SQUARE ROOT OF T TO THE 5TH 
WOULD BE T SQUARED
x THE SQUARE ROOT OF T DIVIDED 
BY, AGAIN, THE SAME THING HERE,
WE HAVE THE SQUARE ROOT OF T TO 
THE 5TH THAT WOULD BE 2T SQUARED
x THE SQUARE ROOT OF T.
IN OUR HOMEWORK THOUGH,
I THINK WE CAN GO AHEAD
AND LEAVE IT IN THIS FORM HERE.
I HOPE YOU FOUND THIS HELPFUL.
