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Popular >

d/(dt)(sqrt(x)ln(t))

Solution

\frac{d}{d t} (x ln (t))

Solution

\frac{ln ( t ) \frac{d x}{d t}}{2 x} + \frac{x}{t}

Solution steps

\frac{d}{d t} (x ln (t))

Apply the Product Rule:

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

f = x, g = ln (t)

= \frac{d}{d t} (x) ln (t) + \frac{d}{d t} (ln (t)) x

\frac{d}{d t} (x) = \frac{\frac{d x}{d t}}{2 x}

\frac{d}{d t} (ln (t)) = \frac{1}{t}

= \frac{\frac{d x}{d t}}{2 x} ln (t) + \frac{1}{t} x

Simplify

\frac{\frac{d x}{d t}}{2 x} ln (t) + \frac{1}{t} x : \frac{ln ( t ) \frac{d x}{d t}}{2 x} + \frac{x}{t}

= \frac{ln ( t ) \frac{d x}{d t}}{2 x} + \frac{x}{t}

Popular Examples

area y=x^3-x,y=3x,-2,2 area

y = x^{3} - x, y = 3 x, - 2, 2

derivative of (1+3x^2(x-x^2))

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

derivative of (-4x+sqrt(1+x)/(x^2-1))

\frac{d}{d x} (\frac{- 4 x + 1 + x}{x ^{2} - 1})

tangent of y=2^{x^2-3x+5},\at x=4 tangent of

y = 2^{x^{2} - 3 x + 5}, at x = 4

integral of 1/(x(x+4)^2)

\int \frac{1}{x ( x + 4 ) ^{2}} d x

Frequently Asked Questions (FAQ)

What is the d/(dt)(sqrt(x)ln(t)) ?
The d/(dt)(sqrt(x)ln(t)) is (ln(t)(dx)/(dt)}{2sqrt(x)}+\frac{sqrt(x))/t
What is the first d/(dt)(sqrt(x)ln(t)) ?
The first d/(dt)(sqrt(x)ln(t)) is (ln(t)(dx)/(dt)}{2sqrt(x)}+\frac{sqrt(x))/t