What is the derivative of (7^2)/y-ln(y)-1 ?

The derivative of (7^2)/y-ln(y)-1 is -(49)/(y^2)-1/y

What is the first derivative of (7^2)/y-ln(y)-1 ?

The first derivative of (7^2)/y-ln(y)-1 is -(49)/(y^2)-1/y

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derivative of

\frac{7 ^{2}}{y} - ln (y) - 1

- \frac{49}{y ^{2}} - \frac{1}{y}

Solution steps

\frac{d}{d y} (\frac{7 ^{2}}{y} - ln (y) - 1)

Simplify

\frac{7 ^{2}}{y} - ln (y) - 1 : \frac{49}{y} - ln (y) - 1

= \frac{d}{d y} (\frac{49}{y} - ln (y) - 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d y} (\frac{49}{y}) - \frac{d}{d y} (ln (y)) - \frac{d}{d y} (1)

\frac{d}{d y} (\frac{49}{y}) = - \frac{49}{y ^{2}}

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

\frac{d}{d y} (1) = 0

= - \frac{49}{y ^{2}} - \frac{1}{y} - 0

Simplify

= - \frac{49}{y ^{2}} - \frac{1}{y}

2 \cdot 5^{x}

\int \frac{10 x}{64 x ^{4} - 49} d x

\frac{d}{d x} (sin (x) tan (x))

((f (x))^{2})^{'}

\frac{d}{d x} (cos (2 x) e^{2 x})

What is the derivative of (7^2)/y-ln(y)-1 ?
The derivative of (7^2)/y-ln(y)-1 is -(49)/(y^2)-1/y
What is the first derivative of (7^2)/y-ln(y)-1 ?
The first derivative of (7^2)/y-ln(y)-1 is -(49)/(y^2)-1/y