00:01
In this problem, we are given that y equals to modulus of x over x raised to the power 8, and we are asked to find out d .y over d x.
00:13
So first let us look at the definition of modulus of x.
00:16
Modulus of x is defined as x when x is greater than equal to 0.
00:20
And it is defined as negative x when x is less than 0.
00:25
So we can look at this as square root of x squared, since square root of x squared is also defined as x when x is greater than or equal to 0, and it is defined as negative x when x is less than 0.
00:41
So this implies that modulus of x equals to square root of x squared.
00:47
So now let us substitute this.
00:49
We have y equals to square root of x squared over x raised to the power 8, which can be written as x squared raised to the power 1 over 2.
01:00
To the whole divided by x raised to the power 8.
01:05
And here, since we have a numerator and a denominator, we use the quotient rule, that is, u over v, the whole prime equals to v u prime minus uv prime, the whole divided by v squared.
01:19
So here we have dy over d x equals to v, which is x raised to the power 8 times u prime, which is 1 over 2 times, 1 over x squared whole power 1 over 2 times the derivative of x squared which is 2x minus now the numerator remains as it is times the derivative of the denominator which is 8 times x raised to the pass 7 the whole divided by x raised to the power 8 the whole square which is x raised to the power 16 so now let us simplify this here 2 and 2 would get cancelled and let us replace x squared raised to the power 1 over 2 in both of these places as modulus x...