y = log tanx differentiate wrt x dy/dx = 1/tanx .d(tanx)/dx = 1/tanx .sec²x = sec²x/tanx [ sec²x = 1 + tan²x use this ]

2017年8月18日 · What does $ d\tan(x) = \sec^2(x)\,dx$ mean? I've seen it used in integration problems to make it more simpler.

2022年2月6日 · I couldn't prove the derivative of $\tan x$ by using the infinitesimal approach to the derivatives. Here's my solution, but I stuck at the last step: \begin{align} \tag{1}\frac{\mathrm{d} }{\mathrm...

#d/dx (tanx) = sec^2(x) #, so: #int sec^5x dx = int sec^2(x) sec^3(x)dx# #int sec^5x dx = int sec^3(x)d(tanx)# #int sec^5x dx = tanxsec^3x - int tanx d(sec^3(x))# and as: #d/dx (sec^3(x)) = 3sec^2(x) d/dx sec(x) = 3sec^3(x) tanx# we have: #int sec^5x dx = tanxsec^3x - 3int tan^2x sec^3x dx# use now the trigonometric identity:

2016年3月28日 · If we rewrite $\\displaystyle \\frac {d} {dx} \\cot(x)$ as $\\displaystyle \\frac {d} {dx} \\frac {1} {\\tan(x)}$ and then apply the quotient rule, we get to ...

Your approach is correct, but here's another way to get the same answer. Notice that both the derivative (i.e., $3 \tan^2 x \sec^2 x - 3 \sec^2 x + 3$) and the right hand side ($=3 \tan^4 x$) involve only $\tan^2x$ terms.

Here is a straightforward way of finding the derivative manipulating only sines and cosines: $$ \frac{d}{dx} \left[ \frac{\sec(x)}{1+\tan(x)} \right] \\ = \frac{d}{dx ...

2017年5月4日 · However, there is a nice geometric proof that $\frac {d}{dx} \tan x = \sec^2 x$ that is worth knowing. In the figure we have two right triangles. The base is 1.

2019年7月24日 · I understand why the derivative of $\\sin x$ is $\\cos x$, and why the derivative of $\\cos x$ is $-\\sin x$ in a geometric way. But I can not understand why the derivative of $\\tan x$ is $\\sec^2 x$....

Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$ How should I proceed? I tried: Then another which I think is wrong: What should I be doing?