00:01
In this problem we are provided that y equals to u power 5 and u equals to x squared plus 2 and we are asked to find out d y over du du du over d x and d y over d x let us start with d y over d u.
00:21
D for d -y over d -u.
00:33
Next, moving towards d -u over d -x, differentiating u with respect to x, we get 2 times x plus 0 since the derivative of a constant is 0.
00:45
So this equals to 2x.
00:47
Therefore, this is the final answer for d -u over d -x.
00:52
Next, for calculating d -y over d -x, we use the chain rule as d -y over -d -u times d -d -u.
01:01
Over d x.
01:03
So let us substitute the values we get 5 times u power 4 multiplied with 2 times x.
01:11
Substituting the value of u we get 5 times x squared plus 2 whole power 4 times 2 x which simplifies to 10 times x multiplied with x squared plus 2 whole power 4.
01:28
So therefore this is our final answer for for, d -y over d -x.
01:35
Next, moving towards sub -part b, in sub -part b we are provided that y equals to u -par -4 and u equals to 8 times x squared minus x plus 7.
01:50
Here again, we are asked to calculate d -y over d -u, d -u over d -x and d -y over d -x.
01:56
Let us start with d -y over d -d -u...