What is the derivative of 0.1(0.7n^2+9n+29)^{(1/2)} ?

The derivative of 0.1(0.7n^2+9n+29)^{(1/2)} is (0.05(1.4n+9))/((0.7n^2+9n+29)^{1/2)}

What is the first derivative of 0.1(0.7n^2+9n+29)^{(1/2)} ?

The first derivative of 0.1(0.7n^2+9n+29)^{(1/2)} is (0.05(1.4n+9))/((0.7n^2+9n+29)^{1/2)}

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derivative of

0.1 (0.7 n^{2} + 9 n + 29)^{(\frac{1}{2})}

\frac{0.05 ( 1.4 n + 9 )}{( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}}

Solution steps

\frac{d}{d n} (0.1 (0.7 n^{2} + 9 n + 29)^{\frac{1}{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.1 \frac{d}{d n} ((0.7 n^{2} + 9 n + 29)^{\frac{1}{2}})

Apply the chain rule:

\frac{1}{2 ( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}} \frac{d}{d n} (0.7 n^{2} + 9 n + 29)

= \frac{1}{2 ( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}} \frac{d}{d n} (0.7 n^{2} + 9 n + 29)

\frac{d}{d n} (0.7 n^{2} + 9 n + 29) = 1.4 n + 9

= 0.1 \cdot \frac{1}{2 ( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}} (1.4 n + 9)

Simplify

0.1 \cdot \frac{1}{2 ( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}} (1.4 n + 9) : \frac{0.05 ( 1.4 n + 9 )}{( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}}

= \frac{0.05 ( 1.4 n + 9 )}{( 0.7 n ^{2} + 9 n + 29 ) ^{\frac{1}{2}}}

\frac{d}{d x} (\frac{5}{x})

\frac{89}{2} s

(e^{- \frac{1}{x}})^{^{'}}

f (x) = (x^{2} + 2 x)^{3} (7 x + 1)^{4}

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

What is the derivative of 0.1(0.7n^2+9n+29)^{(1/2)} ?
The derivative of 0.1(0.7n^2+9n+29)^{(1/2)} is (0.05(1.4n+9))/((0.7n^2+9n+29)^{1/2)}
What is the first derivative of 0.1(0.7n^2+9n+29)^{(1/2)} ?
The first derivative of 0.1(0.7n^2+9n+29)^{(1/2)} is (0.05(1.4n+9))/((0.7n^2+9n+29)^{1/2)}