Hello dear friends. In this lesson, we will tell you step by step what is the derivative of x.lnx.
First, let's write the product derivative formula:
Firstl, we can let $y=f(x).g(x)$
So derivate of $y’=f(x)’.g(x)+g(x)’.f(x)$
Now, we can go our solution, derivative of x.ln(x)
So, while taking the derivative of x.lnx, we take the derivative of the 1st and multiply it by the 2nd. Then we take the derivative of the 2nd and multiply it by the 1st. Then we add the two. Thus, we find its derivative. Now let's do it step by step.
Let $y=x.ln(x)$
x.lnx Derivative:
1-) $y’=x'.ln(x)+ln'(x).x$
2-) $y'=1.ln(x)+\frac{1}{x}.x$
3-) $y'=ln(x)+1$
4-) As a result, our answer is lnx+1
x.lnx derivative = ln(x)+1
Thank you so much...
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