What is the derivative of y=x^{3/2}(x+ce^x) ?

The derivative of y=x^{3/2}(x+ce^x) is (3ce^xx^{1/2}+2ce^xx^{3/2}+5x^{3/2})/2

What is the first derivative of y=x^{3/2}(x+ce^x) ?

The first derivative of y=x^{3/2}(x+ce^x) is (3ce^xx^{1/2}+2ce^xx^{3/2}+5x^{3/2})/2

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derivative of

y = x^{\frac{3}{2}} (x + c e^{x})

\frac{3 c e ^{x} x ^{\frac{1}{2}} + 2 c e ^{x} x ^{\frac{3}{2}} + 5 x ^{\frac{3}{2}}}{2}

Solution steps

\frac{d}{d x} (x^{\frac{3}{2}} (x + c e^{x}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{\frac{3}{2}}, g = x + c e^{x}

= \frac{d}{d x} (x^{\frac{3}{2}}) (x + c e^{x}) + \frac{d}{d x} (x + c e^{x}) x^{\frac{3}{2}}

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

\frac{d}{d x} (x + c e^{x}) = 1 + c e^{x}

= \frac{3 x ^{\frac{1}{2}}}{2} (x + c e^{x}) + (1 + c e^{x}) x^{\frac{3}{2}}

Simplify

\frac{3 x ^{\frac{1}{2}}}{2} (x + c e^{x}) + (1 + c e^{x}) x^{\frac{3}{2}} : \frac{3 c e ^{x} x ^{\frac{1}{2}} + 2 c e ^{x} x ^{\frac{3}{2}} + 5 x ^{\frac{3}{2}}}{2}

= \frac{3 c e ^{x} x ^{\frac{1}{2}} + 2 c e ^{x} x ^{\frac{3}{2}} + 5 x ^{\frac{3}{2}}}{2}

\int 10 - (14 - 2 (16 - 2 x)^{\frac{1}{2}}) d x

\int x^{4} - 8 x^{2} + 16 d x

x \to - 3 lim (\frac{x}{( x + 3 ) ^{4}})

y^{^{''}} - k y = k x e^{- x}

\frac{d}{d x} (42)

What is the derivative of y=x^{3/2}(x+ce^x) ?
The derivative of y=x^{3/2}(x+ce^x) is (3ce^xx^{1/2}+2ce^xx^{3/2}+5x^{3/2})/2
What is the first derivative of y=x^{3/2}(x+ce^x) ?
The first derivative of y=x^{3/2}(x+ce^x) is (3ce^xx^{1/2}+2ce^xx^{3/2}+5x^{3/2})/2