Y equal to X square + 2 raise to x find DY by DX
Answers
Answer: a calculus tutorial on how to find the first derivative of y = x x for x > 0
Explanation:
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Note that the function defined by y = x x is neither a power function of the form x k nor an exponential function of the form b x and the formulas of Differentiation of these functions cannot be used. We need to find another method to find the first derivative of the above function.
If y = x x and x > 0 then ln y = ln (x x)
Use properties of logarithmic functions to expand the right side of the above equation as follows.
ln y = x ln x
We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right.
y '(1 / y) = ln x + x(1 / x) = ln x + 1 , where y ' = dy/dx
Multiply both sides by y
y ' = (ln x + 1)y
Substitute y by x x to obtain
y ' = (ln x + 1)x x
Exercise: Find the first derivative of y = xx - 2
Answer to Above Exercise: y ' = x x - 3 (x ln x + x - 2)
More on differentiation and derivatives
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