What is the derivative of y=((sqrt(x))/7+9)^{3/2} ?

The derivative of y=((sqrt(x))/7+9)^{3/2} is (3(sqrt(x)+63)^{1/2})/(28*7^{1/2)x^{1/2}}

What is the first derivative of y=((sqrt(x))/7+9)^{3/2} ?

The first derivative of y=((sqrt(x))/7+9)^{3/2} is (3(sqrt(x)+63)^{1/2})/(28*7^{1/2)x^{1/2}}

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derivative of

y = (\frac{x}{7} + 9)^{\frac{3}{2}}

\frac{3 ( x + 63 ) ^{\frac{1}{2}}}{28 \cdot 7 ^{\frac{1}{2}} x ^{\frac{1}{2}}}

Solution steps

\frac{d}{d x} ((\frac{x}{7} + 9)^{\frac{3}{2}})

Simplify

(\frac{x}{7} + 9)^{\frac{3}{2}} : (\frac{x + 63}{7})^{\frac{3}{2}}

= \frac{d}{d x} ((\frac{x + 63}{7})^{\frac{3}{2}})

Apply the chain rule:

\frac{3 ( \frac{x + 63}{7} ) ^{\frac{1}{2}}}{2} \frac{d}{d x} (\frac{x + 63}{7})

= \frac{3 ( \frac{x + 63}{7} ) ^{\frac{1}{2}}}{2} \frac{d}{d x} (\frac{x + 63}{7})

\frac{d}{d x} (\frac{x + 63}{7}) = \frac{1}{14 x ^{\frac{1}{2}}}

= \frac{3 ( \frac{x + 63}{7} ) ^{\frac{1}{2}}}{2} \cdot \frac{1}{14 x ^{\frac{1}{2}}}

Simplify

\frac{3 ( \frac{x + 63}{7} ) ^{\frac{1}{2}}}{2} \cdot \frac{1}{14 x ^{\frac{1}{2}}} : \frac{3 ( x + 63 ) ^{\frac{1}{2}}}{28 \cdot 7 ^{\frac{1}{2}} x ^{\frac{1}{2}}}

= \frac{3 ( x + 63 ) ^{\frac{1}{2}}}{28 \cdot 7 ^{\frac{1}{2}} x ^{\frac{1}{2}}}

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

\frac{d y}{d x} = x + x y

f (x) = 2 ln (x)

\frac{- s - 1}{s ^{2} + s + 1}

y = e^{x}, y = e^{- x}, [0, 1]

What is the derivative of y=((sqrt(x))/7+9)^{3/2} ?
The derivative of y=((sqrt(x))/7+9)^{3/2} is (3(sqrt(x)+63)^{1/2})/(28*7^{1/2)x^{1/2}}
What is the first derivative of y=((sqrt(x))/7+9)^{3/2} ?
The first derivative of y=((sqrt(x))/7+9)^{3/2} is (3(sqrt(x)+63)^{1/2})/(28*7^{1/2)x^{1/2}}