# How do you find the derivative for #(t^1.7 + 8)/( t^1.4 + 6)#?

You can use the *quotient rule*.

In your case, you have

and

This is equivalent to

Finally, you can write

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To find the derivative of the function ( \frac{t^{1.7} + 8}{t^{1.4} + 6} ), you can use the quotient rule, which states that the derivative of ( \frac{f(x)}{g(x)} ) is ( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ). Applying this rule, the derivative of the given function is:

[ \frac{(1.7t^{0.7})(t^{1.4} + 6) - (t^{1.7} + 8)(1.4t^{0.4})}{(t^{1.4} + 6)^2} ]

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To find the derivative of the given function (\frac{t^{1.7} + 8}{t^{1.4} + 6}), you can use the quotient rule, which states that if you have a function of the form (\frac{f(x)}{g(x)}), the derivative is given by:

[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{(g(x))^2} ]

Apply this rule to find the derivative of the given function. You'll need to find the derivatives of (f(t) = t^{1.7}) and (g(t) = t^{1.4}) using the power rule, and then apply the quotient rule.

So, the derivative ( \frac{d}{dt} \left( \frac{t^{1.7} + 8}{t^{1.4} + 6} \right) ) can be computed using the quotient rule.

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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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