Question

differentiate the function.
$f(r)=\frac{6}{r^{3}}$
$f(r)=

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differentiate the function.
$h(t)=sqrt4{t}-4e^{t}$
$h(t)=$

Explanation:

Step1: Rewrite the function

Rewrite $F(r)=\frac{6}{r^{3}}$ as $F(r) = 6r^{- 3}$ using the rule $\frac{1}{x^{n}}=x^{-n}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $F(r)=6r^{-3}$, we have $a = 6$ and $n=-3$. So $F^\prime(r)=6\times(-3)r^{-3 - 1}$.

Step3: Simplify the result

$F^\prime(r)=-18r^{-4}=-\frac{18}{r^{4}}$.

Step1: Differentiate each term separately

The derivative of $y = t^{\frac{1}{4}}$ using the power - rule ($y = ax^{n}$, $y^\prime=anx^{n - 1}$) with $a = 1$ and $n=\frac{1}{4}$ is $y^\prime=\frac{1}{4}t^{\frac{1}{4}-1}=\frac{1}{4}t^{-\frac{3}{4}}$. The derivative of $y=-4e^{t}$ using the rule that the derivative of $e^{t}$ is $e^{t}$ is $y^\prime=-4e^{t}$.

Step2: Combine the derivatives

$h^\prime(t)=\frac{1}{4}t^{-\frac{3}{4}}-4e^{t}=\frac{1}{4t^{\frac{3}{4}}}-4e^{t}$.

Answer:

$-\frac{18}{r^{4}}$

Now for $h(t)=\sqrt[4]{t}-4e^{t}=t^{\frac{1}{4}}-4e^{t}$: