What is the d/(dt)(1/3 (2k+t^2)^{3/2}) ?

The d/(dt)(1/3 (2k+t^2)^{3/2}) is t(2k+t^2)^{1/2}

What is the first d/(dt)(1/3 (2k+t^2)^{3/2}) ?

The first d/(dt)(1/3 (2k+t^2)^{3/2}) is t(2k+t^2)^{1/2}

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\frac{d}{d t} (\frac{1}{3} (2 k + t^{2})^{\frac{3}{2}})

t (2 k + t^{2})^{\frac{1}{2}}

Solution steps

\frac{d}{d t} (\frac{1}{3} (2 k + t^{2})^{\frac{3}{2}})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d t} ((2 k + t^{2})^{\frac{3}{2}})

Apply the chain rule:

\frac{3 ( 2 k + t ^{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d t} (2 k + t^{2})

= \frac{3 ( 2 k + t ^{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d t} (2 k + t^{2})

\frac{d}{d t} (2 k + t^{2}) = 2 t

= \frac{1}{3} \cdot \frac{3 ( 2 k + t ^{2} ) ^{\frac{1}{2}}}{2} \cdot 2 t

Simplify

\frac{1}{3} \cdot \frac{3 ( 2 k + t ^{2} ) ^{\frac{1}{2}}}{2} \cdot 2 t : t (2 k + t^{2})^{\frac{1}{2}}

= t (2 k + t^{2})^{\frac{1}{2}}

\int \frac{tan ^{2} ( x )}{x} d x

\frac{d}{d x} (\frac{3 cos ( x )}{x ^{4} - 2 x + 2})

f (θ) = sec (θ)

- 0.02 x^{3} + 0.37 x^{2} + 241 x + 17905

(\frac{t}{t + 1 ( t )})^{^{'}}

What is the d/(dt)(1/3 (2k+t^2)^{3/2}) ?
The d/(dt)(1/3 (2k+t^2)^{3/2}) is t(2k+t^2)^{1/2}
What is the first d/(dt)(1/3 (2k+t^2)^{3/2}) ?
The first d/(dt)(1/3 (2k+t^2)^{3/2}) is t(2k+t^2)^{1/2}