How do you use the definition of a derivative to find the derivative of #g(t) = 7/sqrt(t)#?

Answer 1

# =7/(2 sqrt (t^3))#

#g(t) = 7 / sqrt t#
by definition #g'(t) =lim_(h to 0) (g(t + h) - g(t))/(h)#
# =lim_(h to 0) 1/h * (7 / sqrt (t + h) - 7 / sqrt t)#
lifting the constant term out # = 7 lim_(h to 0) 1/h * (1 / sqrt (t + h) - 1 / sqrt t)#
combining the fractions # =7 lim_(h to 0) 1/h * (sqrt (t ) - sqrt (t+h))/(sqrt (t+h)sqrt (t))#
# =7 lim_(h to 0) 1/h * (sqrt (t ) - color(red)(sqrt(t)) sqrt (color(blue)(1+h/t)))/(sqrt (t+h)sqrt (t))#
cancelling #sqrt t# # =7 lim_(h to 0) 1/h * (1 - sqrt (1+h/t))/(sqrt (t+h))#

Now, by generalised Binomial Theorem:

#sqrt (1+h/t) = (1+h/t) ^(1/2) = 1 + 1/2 h/t + O(h^2)#
# implies 7 lim_(h to 0) 1/h * (1 - (1 + 1/2 h/t + O(h^2)))/(sqrt (t+h))#
# =7 lim_(h to 0) 1/h * ( - 1/2 h/t + O(h^2))/(sqrt (t+h))#
# =7 lim_(h to 0) ( - 1/(2t) + O(h))/(sqrt (t+h))#
the limit of the quotient is the quotient of the limits where the limits are known # =7 (lim_(h to 0) - 1/(2t) + O(h))/(lim_(h to 0) sqrt (t+h))#
# =7 ( - 1/(2t) )/( sqrt (t))#
# =7/(2 sqrt (t^3))#
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Answer 2

To find the derivative of ( g(t) = \frac{7}{\sqrt{t}} ), we use the definition of a derivative, which states that the derivative of a function represents the rate of change of that function with respect to its independent variable.

Using the definition of the derivative, we have:

[ g'(t) = \lim_{{h \to 0}} \frac{g(t + h) - g(t)}{h} ]

Substituting ( g(t) = \frac{7}{\sqrt{t}} ), we get:

[ g'(t) = \lim_{{h \to 0}} \frac{\frac{7}{\sqrt{t+h}} - \frac{7}{\sqrt{t}}}{h} ]

Now, we'll simplify the expression:

[ g'(t) = \lim_{{h \to 0}} \frac{7\sqrt{t} - 7\sqrt{t+h}}{h\sqrt{t}\sqrt{t+h}} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7\sqrt{t} - 7\sqrt{t+h}}{h\sqrt{t}\sqrt{t+h}} \times \frac{\sqrt{t} + \sqrt{t+h}}{\sqrt{t} + \sqrt{t+h}} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7t - 7(t+h)}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7t - 7t - 7h}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{-7h}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{-7}{\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

Now, plugging in ( h = 0 ), we get:

[ g'(t) = \frac{-7}{2t^{3/2}} ]

Therefore, the derivative of ( g(t) = \frac{7}{\sqrt{t}} ) is ( g'(t) = \frac{-7}{2t^{3/2}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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