If y=xn log x+x(log x)n , then (dy/dx) is equal to

Question

If y=xn log x+x(log x)n , then (dy/dx) is equal to (A)  xn-1(1+n log x)+(log x)n-1[n+log x]  (B)  xn-2(1+n log x)+(log x)n-1[n+log x]  (C)  xn-1(1+n log x)+(log x)n-1[n-log x]  (D) None of the above

Tardigrade Admin · Accepted Answer

Given, y=xn log x+x( log x)n (d y/d x)=n xn-1 log x+xn ⋅ (1/x)+x n( log x)n-1((1/x)) +1 ⋅( log x)n =xn-1(1+n log x)+( log x)n-1[n+ log x]

Q. If $y = x^{n} l o g x + x (l o g x)^{n}$ , then $\frac{d y}{d x}$ is equal to

$x^{n - 1} (1 + n l o g x) + (l o g x)^{n - 1} [n + l o g x]$

$x^{n - 2} (1 + n l o g x) + (l o g x)^{n - 1} [n + l o g x]$

$x^{n - 1} (1 + n l o g x) + (l o g x)^{n - 1} [n - l o g x]$

None of the above

Given, $y = x^{n} lo g x + x (lo g x)^{n}$
$\frac{d y}{d x} = n x^{n - 1} lo g x + x^{n} \cdot \frac{1}{x} + x n (lo g x)^{n - 1} (\frac{1}{x}) < b r / >< b r / > + 1 \cdot (lo g x)^{n}$
$= x^{n - 1} (1 + n lo g x) + (lo g x)^{n - 1} [n + lo g x]$

Q. If y=xnlogx+x(logx)n , then dxdy​ is equal to

xn−1(1+nlogx)+(logx)n−1[n+logx]

xn−2(1+nlogx)+(logx)n−1[n+logx]

xn−1(1+nlogx)+(logx)n−1[n−logx]