00:01
In this question, we need to differentiate our equations which are given here.
00:06
Differentiate.
00:09
So our first equation is given as f of y is equal to 1 upon y squared minus 7 by y to 5 % by 5 to the bar of 4 multiplied by y plus 3y cube.
00:25
So when we differentiate this equation first time, our first derivative f dash 5 is equal to t by d by of our equation which is 1 upon y square minus 7 by y to the power of 4 and now as multiplied by y plus 3 y q that is equal to as applying the product rule here on applying the product rule we will get here f multiplied by g whole dash is equal to f dash g and it by f dash g so therefore our value for f will be equal to 1 upon y square minus 1 upon y to the part of 4 and we will take g as y plus 3 y cube so therefore our solution will be as d by d ,y for 1 upon y squared minus 7 upon y to the power of 4, y plus 3 by q added by d by d by d by now, y plus 3 by q multiplied by 1 upon y square minus 1 upon y to the power of 4.
02:12
So that is equal to minus 2, y by 3 plus 28 y by 5 by 4.
02:19
5 multiplied by y plus 3 y cube and it by mnix will be as 1 plus 9 y square 1 upon by square minus 1 of 1 y to the power of 4 i'm sorry this is 7 upon sorry for this mistake 7 upon y to the part of 4 and when we simplify we will get here 20 by y square plus 21 by 4 to the power of 4 added by 3.
02:58
So this is our solution for the f -dash 5.
03:03
So we have here our first derivative of the given equation.
03:08
This will be our answer for first version.
03:12
Now let's move on to the second part of the question.
03:14
So in question number 2, our equation is equal to e to the power of p multiplied by p plus p under root p.
03:26
This is an equation so its first derivative will be as a to the bar of or we can say d by d by to the whole value e to the part of p p plus p square root p that is e to the part of p d by d p within the bracket we have p plus p square root p added by p plus p square root p d by d d p e to the bar of p and now it is e to the bar of p one plus three by two square root of p added by e to the bar of p one p plus p square root of p so therefore are first derivative of this equation will be, e to the bar of p will be taken as common.
04:38
It will be as 1 plus p plus p square root p plus 3 by 2 square root p.
04:49
So here we have our second solution.
04:52
This is our first derivative of the given equation.
04:56
Answer for part b...