Question

Find t′(x) from t(x)=√(−3x−7) using the definition of a derivative.

Answer 1

By definition of the derivative,

`\displaystyle t'(x) = \lim_(h\to0) \frac{t(x+h) - t(x)}h`

`\displaystyle t'(x) = \lim_(h\to0) \frac{√(-3(x+h)-7) - √(-3x-7)}h`

Rationalize the numerator by multiplying the fraction uniformly by its conjugate:

`\displaystyle t'(x) = \lim_(h\to0) \frac{√(-3(x+h)-7) - √(-3x-7)}h * (√(-3(x+h)-7) + √(-3x - 7))/(√(-3(x+h)-7) + √(-3x - 7))`

`\displaystyle t'(x) = \lim_(h\to0) (\left(√(-3(x+h)-7)\right)^2 - \left(√(-3x-7)\right)^2)/(h \left(√(-3(x+h)-7) + √(-3x - 7)\right))`

`\displaystyle t'(x) = \lim_(h\to0) ((-3(x+h)-7) - (-3x-7))/(h \left(√(-3(x+h)-7) + √(-3x - 7)\right))`

`\displaystyle t'(x) = \lim_(h\to0) (-3h)/(h \left(√(-3(x+h)-7) + √(-3x - 7)\right))`

`\displaystyle t'(x) = -3 \lim_(h\to0) \frac1{√(-3(x+h)-7) + √(-3x - 7)}`

The remaining limand is continuous at h = 0, so we can substitute h = 0 directly and get a limit/derivative of

`\displaystyle t'(x) = -\frac3{√(-3(x+0)-7) + √(-3x - 7)} = \boxed{-\frac3{2√(-3x-7)}}`