What is the derivative of (t^4+t^5-1)/(t^6) ?

The derivative of (t^4+t^5-1)/(t^6) is (-t^5-2t^4+6)/(t^7)

What is the first derivative of (t^4+t^5-1)/(t^6) ?

The first derivative of (t^4+t^5-1)/(t^6) is (-t^5-2t^4+6)/(t^7)

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derivative of

\frac{t ^{4} + t ^{5} - 1}{t ^{6}}

\frac{- t ^{5} - 2 t ^{4} + 6}{t ^{7}}

Solution steps

\frac{d}{d t} (\frac{t ^{4} + t ^{5} - 1}{t ^{6}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( t ^{4} + t ^{5} - 1 ) t ^{6} - \frac{d}{d t} ( t ^{6} ) ( t ^{4} + t ^{5} - 1 )}{( t ^{6} ) ^{2}}

\frac{d}{d t} (t^{4} + t^{5} - 1) = 5 t^{4} + 4 t^{3}

\frac{d}{d t} (t^{6}) = 6 t^{5}

= \frac{( 5 t ^{4} + 4 t ^{3} ) t ^{6} - 6 t ^{5} ( t ^{4} + t ^{5} - 1 )}{( t ^{6} ) ^{2}}

Simplify

\frac{( 5 t ^{4} + 4 t ^{3} ) t ^{6} - 6 t ^{5} ( t ^{4} + t ^{5} - 1 )}{( t ^{6} ) ^{2}} : \frac{- t ^{5} - 2 t ^{4} + 6}{t ^{7}}

= \frac{- t ^{5} - 2 t ^{4} + 6}{t ^{7}}

\int (4 x + 1)^{6} d x

y^{3} - (18 x + 8) + 3 x y^{2} y^{^{'}} = 0

\int \frac{1}{y ^{3} y ^{2} - 9} d y

\frac{d}{d θ} (sin (4 θ))

\int x (x^{2} - 2) d x

What is the derivative of (t^4+t^5-1)/(t^6) ?
The derivative of (t^4+t^5-1)/(t^6) is (-t^5-2t^4+6)/(t^7)
What is the first derivative of (t^4+t^5-1)/(t^6) ?
The first derivative of (t^4+t^5-1)/(t^6) is (-t^5-2t^4+6)/(t^7)