What is the d/(dy)(yln(y)-e^{-xy}) ?

The d/(dy)(yln(y)-e^{-xy}) is ln(y)+1+e^{-yx}x

What is the first d/(dy)(yln(y)-e^{-xy}) ?

The first d/(dy)(yln(y)-e^{-xy}) is ln(y)+1+e^{-yx}x

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\frac{d}{d y} (y ln (y) - e^{- x y})

ln (y) + 1 + e^{- y x} x

Solution steps

\frac{d}{d y} (y ln (y) - e^{- x y})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d y} (y ln (y)) - \frac{d}{d y} (e^{- x y})

\frac{d}{d y} (y ln (y)) = ln (y) + 1

\frac{d}{d y} (e^{- x y}) = - x e^{- x y}

= ln (y) + 1 - (- x e^{- x y})

Simplify

= ln (y) + 1 + e^{- y x} x

\int \frac{2 sin ( \frac{1}{5 x} )}{5 x ^{2}} d x

y = \frac{64}{16 + x ^{2}}, y = 1

\int 2 sec (4 x) d x

\frac{1}{s + 42} - \frac{1}{( s + 3 ) ^{4}}

\int - π cos (1 π) t d t

What is the d/(dy)(yln(y)-e^{-xy}) ?
The d/(dy)(yln(y)-e^{-xy}) is ln(y)+1+e^{-yx}x
What is the first d/(dy)(yln(y)-e^{-xy}) ?
The first d/(dy)(yln(y)-e^{-xy}) is ln(y)+1+e^{-yx}x