What is derivative of g(x)=(1-3t)/(6+t) ?

The answer to derivative of g(x)=(1-3t)/(6+t) is (dt)/(dx)=0

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derivative of g(x)=(1-3t)/(6+t)

derivative of

g (x) = \frac{1 - 3 t}{6 + t}

\frac{d t}{d x} = 0

Solution steps

g (x) = \frac{1 - 3 t}{6 + t}

Treat

t

t (x)

Differentiate both sides:

0 = - \frac{19 \frac{d t}{d x}}{( 6 + t ) ^{2}}

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = 0

\frac{d t}{d x} = 0

\frac{d y}{d x} + 4 x^{3} y = x^{3}

\int \frac{41}{x ^{3} - 125} d x

\int \frac{ln ( 15 x )}{x} d x

- 4 \cdot arcsin (x)

\int \frac{x ^{3} + 2 x ^{2} - 4}{x ^{2}} d x

What is derivative of g(x)=(1-3t)/(6+t) ?
The answer to derivative of g(x)=(1-3t)/(6+t) is (dt)/(dx)=0