What is derivative of f(x)= 1/((t^2+3t+1)^{5/2)} ?

The answer to derivative of f(x)= 1/((t^2+3t+1)^{5/2)} is (dt)/(dx)=0

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derivative of f(x)= 1/((t^2+3t+1)^{5/2)}

derivative of

f (x) = \frac{1}{( t ^{2} + 3 t + 1 ) ^{\frac{5}{2}}}

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

Solution steps

f (x) = \frac{1}{( t ^{2} + 3 t + 1 ) ^{\frac{5}{2}}}

Treat

t

t (x)

Differentiate both sides:

0 = - \frac{5 ( 2 t \frac{d t}{d x} + 3 \frac{d t}{d x} )}{2 ( t ^{2} + 3 t + 1 ) ^{\frac{7}{2}}}

Isolate

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

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

x y^{^{'}} + y = 6 x y^{2}, y (1) = 9

f (t) = 6 t sin (π t)

(\frac{y}{x} + 9 x) d x + (ln (x) - 4) d y = 0

\frac{\partial}{\partial x} (x^{2} + 3 x y + y^{2})

y^{^{''}} - 16 y^{^{'}} + 64 y = 0

What is derivative of f(x)= 1/((t^2+3t+1)^{5/2)} ?
The answer to derivative of f(x)= 1/((t^2+3t+1)^{5/2)} is (dt)/(dx)=0