We want to solve the differential quotient:
\(\lim_{h \to 0} \frac{tan(3*(x + h)) - tan(3x)}{h}\)
The solution is:
\(3*sec(3x)^2\)
We start by expanding the first term in the numerator.
\(\lim_{h \to 0} \frac{tan(3*x + 3*h) - tan(3x)}{h}\)
We can use the rule:
\(Tan(3x + 3h) = \frac{tan(3x) + tan(3h)}{1 - tan(3x)*tan(3h)}\)
And replace that in the numerator, so we get:
\(\lim_{h \to 0} \frac{\frac{tan(3x) + tan(3h)}{1 - tan(3x)*tan(3h)} - tan(3x)}{h}\\\\\lim_{h \to 0} \frac{\tan(3x) + tan(3h) - tan(3x)*(1 - tan(3x)*tan(3h))}{h*(1 - tan(3x)*tan(3h))}\\\\\lim_{h \to 0} \frac{ tan(3h) - tan(3x)^2*tan(3h)}{h*(1 - tan(3x)*tan(3h))}\\\\\lim_{h \to 0} \frac{ tan(3h)*(1 - tan(3x)^2)}{h*(1 - tan(3x)*tan(3h))}\)
And we know that:
\((1 - tan(3x)^2) = sec(3x)^2\)
Then we got:
\(lim_{h \to 0} \frac{ tan(3h)*sec(3x)^2}{h*(1 - tan(3x)*tan(3h))}\)
Now we can rewrite the tangent of 3h as the quotient of a sine and cosine:
\(lim_{h \to 0} \frac{ sin(3h)*sec(3x)^2}{h*(1 - tan(3x)*tan(3h))*cos(3h)}\)
And near zero, we have that:
sin(x) = x
Then knowing that we will take a limit, we can rewrite:
\(lim_{h \to 0} \frac{ 3*h*sec(3x)^2}{h*(1 - tan(3x)*tan(3h))*cos(3h)}\\\\lim_{h \to 0} \frac{ 3*sec(3x)^2}{(1 - tan(3x)*tan(3h))*cos(3h)}\)
Now we just take the limit, cos(30) = 1 and tan(30) = 0, then we get:
\(lim_{h \to 0} \frac{ 3*sec(3x)^2}{(1 - tan(3x)*tan(3h))*cos(3h)} = 3*sec(3x)^2\)
If you want to learn more, you can read:
https://brainly.com/question/20433457
