What is the d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) ?

The d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) is 4{p}^3

What is the first d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) ?

The first d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) is 4{p}^3

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\frac{d}{d p} (p^{4} + a^{6.2} - c^{6.3})

4 p^{3}

Solution steps

\frac{d}{d p} (p^{4} + a^{6.2} - c^{6.3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d p} (p^{4}) + \frac{d}{d p} (a^{6.2}) - \frac{d}{d p} (c^{6.3})

\frac{d}{d p} (p^{4}) = 4 p^{3}

\frac{d}{d p} (a^{6.2}) = 0

\frac{d}{d p} (c^{6.3}) = 0

= 4 p^{3} + 0 - 0

Simplify

= 4 p^{3}

What is the d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) ?
The d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) is 4{p}^3
What is the first d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) ?
The first d/(d{p)}({p}^4+a^{6.2}-c^{6.3}) is 4{p}^3