What is the derivative of (9e^xx^{7/2})/2+e^xx^{9/2} ?

The derivative of (9e^xx^{7/2})/2+e^xx^{9/2} is (36e^xx^{7/2}+63e^xx^{5/2}+4e^xx^{9/2})/4

What is the first derivative of (9e^xx^{7/2})/2+e^xx^{9/2} ?

The first derivative of (9e^xx^{7/2})/2+e^xx^{9/2} is (36e^xx^{7/2}+63e^xx^{5/2}+4e^xx^{9/2})/4

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derivative of

\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}}

\frac{36 e ^{x} x ^{\frac{7}{2}} + 63 e ^{x} x ^{\frac{5}{2}} + 4 e ^{x} x ^{\frac{9}{2}}}{4}

Solution steps

\frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2}) + \frac{d}{d x} (e^{x} x^{\frac{9}{2}})

\frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2}) = \frac{9}{2} (e^{x} x^{\frac{7}{2}} + \frac{7 e ^{x} x ^{\frac{5}{2}}}{2})

\frac{d}{d x} (e^{x} x^{\frac{9}{2}}) = e^{x} x^{\frac{9}{2}} + \frac{9 e ^{x} x ^{\frac{7}{2}}}{2}

= \frac{9}{2} (e^{x} x^{\frac{7}{2}} + \frac{7 e ^{x} x ^{\frac{5}{2}}}{2}) + e^{x} x^{\frac{9}{2}} + \frac{9 e ^{x} x ^{\frac{7}{2}}}{2}

Simplify

\frac{9}{2} (e^{x} x^{\frac{7}{2}} + \frac{7 e ^{x} x ^{\frac{5}{2}}}{2}) + e^{x} x^{\frac{9}{2}} + \frac{9 e ^{x} x ^{\frac{7}{2}}}{2} : \frac{36 e ^{x} x ^{\frac{7}{2}} + 63 e ^{x} x ^{\frac{5}{2}} + 4 e ^{x} x ^{\frac{9}{2}}}{4}

= \frac{36 e ^{x} x ^{\frac{7}{2}} + 63 e ^{x} x ^{\frac{5}{2}} + 4 e ^{x} x ^{\frac{9}{2}}}{4}

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What is the derivative of (9e^xx^{7/2})/2+e^xx^{9/2} ?
The derivative of (9e^xx^{7/2})/2+e^xx^{9/2} is (36e^xx^{7/2}+63e^xx^{5/2}+4e^xx^{9/2})/4
What is the first derivative of (9e^xx^{7/2})/2+e^xx^{9/2} ?
The first derivative of (9e^xx^{7/2})/2+e^xx^{9/2} is (36e^xx^{7/2}+63e^xx^{5/2}+4e^xx^{9/2})/4