What is the derivative of (s-1)^n*t ?

The derivative of (s-1)^n*t is nt(s-1)^{n-1}

What is the first derivative of (s-1)^n*t ?

The first derivative of (s-1)^n*t is nt(s-1)^{n-1}

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derivative of

(s - 1)^{n} \cdot t

n t (s - 1)^{n - 1}

Solution steps

\frac{d}{d s} ((s - 1)^{n} t)

Treat

n, t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= t \frac{d}{d s} ((s - 1)^{n})

Apply the chain rule:

n (s - 1)^{n - 1} \frac{d}{d s} (s - 1)

= n (s - 1)^{n - 1} \frac{d}{d s} (s - 1)

\frac{d}{d s} (s - 1) = 1

= t n (s - 1)^{n - 1} \cdot 1

Simplify

= n t (s - 1)^{n - 1}

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

lo g_{10} (sin (x^{2}))

5 sin (x + 1)

\frac{l}{800}

\frac{5 x ^{3}}{4}

What is the derivative of (s-1)^n*t ?
The derivative of (s-1)^n*t is nt(s-1)^{n-1}
What is the first derivative of (s-1)^n*t ?
The first derivative of (s-1)^n*t is nt(s-1)^{n-1}