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

1 Answer
Jul 30, 2015

You can use the quotient rule.

Explanation:

You can differentiate a function #f(x)# that can be written as the quotient of two other functions, #g(x)# and #h(x)#, by using the quotient rule

#color(blue)(d/dx(f(x)) = (g^'(x) * h(x) - g(x) * h^'(x))/[h(x)]^2#, where #h(x)!=0#

In your case, you have

#g(t) = t^1.7 + 8#

and

#h(y) = t^1.4 + 6#

This means that the derivative of your function #f(t) = g(t)/(h(t))# will be

#f^'(t) = (d/(dt)(t^1.7 + 8) * (t^1.4 + 6) - (t^1.7 + 8) * d/(dt)(t^1.4 + 6))/(t^1.4 + 6)^2#

#f^'(t) = (1.7 * t^0.7 * (t^1.4 + 6) - (t^1.7 + 8) * 1.4 * t^0.4)/(t^1.4 + 6)^2#

This is equivalent to

#f^'(t) = (1/10(17 * t^2.1 + 102 * t^0.7 - 14 * t^2.1 - 112 * t^0.4))/(t^1.4 + 6)^2#

#f^'(t) = (3 * t^2.1 + 102 * t^0.7 - 112 * t^0.4)/(10(t^1.4 + 6)^2)#

Finally, you can write

#f^'(t) = color(green)(1/10 * (t^0.4(3 * t^1.7 + 102 * t^0.3 - 112))/(t^1.4 + 6)^2#