What is the derivative of (e^{y^2})/((2y+1)^5) ?

The derivative of (e^{y^2})/((2y+1)^5) is (2e^{y^2}(y(2y+1)-5))/((2y+1)^6)

What is the first derivative of (e^{y^2})/((2y+1)^5) ?

The first derivative of (e^{y^2})/((2y+1)^5) is (2e^{y^2}(y(2y+1)-5))/((2y+1)^6)

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derivative of

\frac{e ^{y^{2}}}{( 2 y + 1 ) ^{5}}

\frac{2 e ^{y^{2}} ( y ( 2 y + 1 ) - 5 )}{( 2 y + 1 ) ^{6}}

Solution steps

\frac{d}{d y} (\frac{e ^{y^{2}}}{( 2 y + 1 ) ^{5}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d y} ( e ^{y^{2}} ) ( 2 y + 1 ) ^{5} - \frac{d}{d y} ( ( 2 y + 1 ) ^{5} ) e ^{y^{2}}}{( ( 2 y + 1 ) ^{5} ) ^{2}}

\frac{d}{d y} (e^{y^{2}}) = e^{y^{2}} \cdot 2 y

\frac{d}{d y} ((2 y + 1)^{5}) = 10 (2 y + 1)^{4}

= \frac{e ^{y^{2}} \cdot 2 y ( 2 y + 1 ) ^{5} - 10 ( 2 y + 1 ) ^{4} e ^{y^{2}}}{( ( 2 y + 1 ) ^{5} ) ^{2}}

Simplify

\frac{e ^{y^{2}} \cdot 2 y ( 2 y + 1 ) ^{5} - 10 ( 2 y + 1 ) ^{4} e ^{y^{2}}}{( ( 2 y + 1 ) ^{5} ) ^{2}} : \frac{2 e ^{y^{2}} ( y ( 2 y + 1 ) - 5 )}{( 2 y + 1 ) ^{6}}

= \frac{2 e ^{y^{2}} ( y ( 2 y + 1 ) - 5 )}{( 2 y + 1 ) ^{6}}

\frac{d}{d x} (\frac{2}{3} x^{\frac{3}{2}} - \frac{1}{2} x^{\frac{1}{2}})

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

\frac{d}{d x} (\frac{1}{4} e^{- x} + e^{x})

f (x) = x x + 1

x \to 0 lim (\frac{7 ^{x} - 1}{x})

What is the derivative of (e^{y^2})/((2y+1)^5) ?
The derivative of (e^{y^2})/((2y+1)^5) is (2e^{y^2}(y(2y+1)-5))/((2y+1)^6)
What is the first derivative of (e^{y^2})/((2y+1)^5) ?
The first derivative of (e^{y^2})/((2y+1)^5) is (2e^{y^2}(y(2y+1)-5))/((2y+1)^6)