MATH SOLVE

4 months ago

Q:
# find the first, second and third derivatives 1-y=5x5+3x4+x

Accepted Solution

A:

First Derivative:
$$\[y = 5x^5 + 3x^4 + x\]$$
$$\[y' = \frac{d}{dx}(5x^5) + \frac{d}{dx}(3x^4) + \frac{d}{dx}(x)\]$$
$$\[y' = 5 \cdot 5x^{5-1} + 3 \cdot 4x^{4-1} + 1\]$$
$$\[y' = 25x^4 + 12x^3 + 1\]$$
Second Derivative:
$$\[y' = 25x^4 + 12x^3 + 1\]$$
$$\[y'' = \frac{d}{dx}(25x^4) + \frac{d}{dx}(12x^3) + \frac{d}{dx}(1)\]$$
$$\[y'' = 25 \cdot 4x^{4-1} + 12 \cdot 3x^{3-1}\]$$
$$\[y'' = 100x^3 + 36x^2\]$$
Third Derivative:
$$\[y'' = 100x^3 + 36x^2\]$$
$$\[y''' = \frac{d}{dx}(100x^3) + \frac{d}{dx}(36x^2)\]$$
$$\[y''' = 100 \cdot 3x^{3-1} + 36 \cdot 2x^{2-1}\]$$
$$\[y''' = 300x^2 + 72x\]$$