Question

evaluate the derivative of log tanx from first principle
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Answer 1

Answer:

• #d/dx tanx = lim_(h->0) (tan(x+h)-tanx)/h# Using the trigonometric formulas for the sum of two angles: #tan(x+h) = sin(x+h)/cos(x+h)# ...
• #tan(x+h) = (tanx+tan(h))/(1-tanx tan(h))# So: ...
• #d/dx tanx = (1+tan^2x ) lim_(h->0) tan(h)/h1/(1-tanx tan(h))# and as:

hope it's help✌

Answer 2

Frist principle formula for derivative of a function $f(x)$ is

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Given,

$f(x)= log(tanx)$

from first principle,

$f(x+h) = log(tan(x+h)\\f(x)=log(tanx)$

By substituting the above values in the formula we get

$f'(x)= \lim_{h \to 0} \frac{log(tan(x+h))-log(tanx)}{h} ---- > (1)$

Now let us assume

$tanx =u$

and

$tan(x+h) = u+k\\k= tan(x+h)-u$

such that when

condition 1): ($h- > 0$) then ($k- > 0$)

Then by substituting the above values in equation $(1)$ we get

$f'(x) = \lim_{ h\to 0} \frac{log(u+k)-log(u)}{h}$

we know that $log(a)-log(b) = log[\frac{a}{b} ]$

$f'(x) = \lim_{h \to 0} \frac{log[\frac{u+k}{u} ]}{h} \\ \\f'(x)= \lim_{h \to 0} \frac{log[\frac{k}{u}+1] }{h}$

Now, applying the condition 1) for the above equation, we get

$f'(x) = \lim_{k \to 0} \frac{log[\frac{k}{u}+1] }{h*\frac{k}{u} }*\frac{k}{u}$

Now, by modifying it according to the limits and variables we get

$f'(x)= [\lim_{k \to 0} \frac{log[\frac{k}{u}+1] }{\frac{k}{u} }*\frac{k}{u}][ \lim_{h \to 0} \frac{1}{h}]$

From basic formula

$\lim_{x \to 0} \frac{log(x+1)}{x} =1$

here, $x = \frac{k}{u}$ then from the formula we get

$f'(x) = [1*\frac{k}{u}][ \lim_{h \to 0} \frac{1}{h}]$

Now, put back all the assumptions in the above equation. Then we get

$f'(x) = \lim_{h \to 0} \frac{k}{u*h} \\ \\f'(x)= \lim_{h \to 0} \frac{tan(x+h)-tanx}{tanx*h}$

from the trigonometric formula of $tanx = \frac{sinx}{cosx}$ we get the equation as

$f'(x) = \lim_{h \to 0} \frac{\frac{sin(x+h)}{cos(x+h)}-\frac{sinx}{cosx} }{tanx*h} \\ \\f'(x) = \lim_{h \to 0} \frac{sin(x+h)cosx-cos(x+h)sinx}{tanx*h*cos(x+h)*cosx}$

since we know that $sin(A-B) = sinAcosB-cosAsinB$. Then the equation will be

$f'(x) = \lim_{h \to 0} \frac{sin(x+h-x)}{tanx*h*cos(x+h)*cosx}\\ \\ f'(x) = \lim_{h \to 0} \frac{sinh}{tanx*h*cos(h+x)*cosx} \\ \\f'(x) = \lim_{h \to 0} \frac{sinh}{h}*\frac{1}{\frac{sinx}{cosx}*cos(x+h)*cosx }$

since we know that form basic limit formula $\lim_{x \to 0} \frac{sinx}{x}=1$ then

$f'(x) = 1* \lim_{h \to 0} \frac{1}{sinx*cos(x+h)}$

now by applying the limit we get

$f'(x)=\frac{1}{sinx*cosx} \\ \\f'(x) = \frac{2}{2*sinx*cosx}$

we know that, $2*sinx*cosx = sin2x$ then we get

$f'(x) = \frac{2}{sin2x}$

from the basic trigonometric relations we know that $\frac{1}{sinx}=cosecx$ we get

$f'(x)= 2cosec(2x)$

∴ The derivative of $log(tanx)$ by using first principle is $2cosec(2x)$

#SPJ3