y= (1+x)/logx find out in double differenciation
Answers
Answer :
The second derivative of is
Explanation :
To find :
The derivative of :
Knowledge required :
- Quotient rule of differentiation :
- Product rule of differentiation :
- Derivative rule for log x :
- Derivative of a constant term is 0.
- Exponent rule of differentiation :
Solution :
By using the quotient rule of differentiation and substituting the values in it, we get : [Here, u = (x + 1) and v = log(x)]
Now let's find out the second Derivative of the equation :
From there,
First let us differentiate the terms :
Differentiation of :
By applying the chain rule of differentiation, we get :
Differentiation of
By using the quotient rule of differentiation and substituting the values in it, we get :
Now by substituting the derivative of log²x , we get :
Now by substituting the values in the equation, we get :
Answer:
The derivative of \sf{y = \dfrac{1 + x}{log(x)}}y=
log(x)
1+x
is \begin{gathered}\sf{\dfrac{1}{log(x)} - \dfrac{(x + 1)}{(x)log^{2}(x)}} \\ \\\end{gathered}
log(x)
1
−
(x)log
2
(x)
(x+1)