Question

Differentiate. g(t)={t- √ {t} t¹/⁷ }

Answer 1

The derivative
`\(g'(t) = 1 - (1)/(2√(t)) \cdot t^(1/7) - (1)/(7)t^(-6/7)\)` signifies the rate of change of the function g(t) concerning t.

The given function is
`\(g(t) = t - √(t) \cdot t^(1/7)\)`. To differentiate g(t) with respect to t, we use the sum and product rules of differentiation.

1. Differentiate the first term (t) with respect to (t):
`\((d)/(dt)(t) = 1\)`.

2. For the second term √t, apply the chain rule:
`\((d)/(dt)(√(t)) = (1)/(2√(t))\)`.

3. For the third term t^(1/7), apply the power rule:
`\((d)/(dt)(t^(1/7)) = (1)/(7)t^(-6/7)\)`.

Now, combining these results using the sum and product rules, the derivative g'(t) is:

`\[g'(t) = 1 - (1)/(2√(t)) \cdot t^(1/7) - (1)/(7)t^(-6/7)\]`

This expression represents the rate of change of the function g(t) with respect to (t).