Question

d/dx (tanx) = sec(x)^2​

Answer 1

d/dx tanx=d/dx sinx/cosx=d/dx (sinx)/cosx +sinx d/dx (1/cosx)

=1+sinx(sinx/cos²x)=1+sin²x/cos²x=1/cos²x=sec²x

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

y = tanx

Also,

y + δy = tan(x + δx)

Hence,

$\tt{\rightarrow\dfrac{\delta y}{\delta x}=\dfrac{tan(x+\delta x)-tanx}{\delta x}}$

Therefore,

$\tt{\rightarrow\dfrac{dy}{dx}=\lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}}$

$\tt{\rightarrow \lim_{\delta x \to 0}\dfrac{tan(x+\delta x)-tanx}{\delta x}}$

$\tt{\rightarrow \lim_{\delta x \to 0}\dfrac{[sin(x+\delta x)/cos(x+\delta x)-sinx/cosx}{\delta x}}$

$\tt{\rightarrow \lim_{\delta x \to 0}\dfrac{[sin(x+\delta x) cosx-cos(x+\delta x)sinx}{\delta x\times cos(x+\delta x)\times cosx}}$

$\tt{\rightarrow \dfrac{1}{cosx}\times \lim_{\delta x \to 0}\dfrac{sin\delta x}{\delta x[cos(x+\delta x)]}}$

$\tt{\rightarrow \dfrac{1}{cosx}\times \lim_{\delta x \to 0}\dfrac{sin\delta x}{\delta x}[\lim_{x \to 0}\dfrac{1}{cos(x+\delta x)}]}$

$\tt{\rightarrow\dfrac{1}{cosx}\times 1\times\dfrac{1}{cosx}}$

$\tt{\rightarrow\dfrac{1}{cos^2 x}=sec^2 x}$

Therefore,

$\tt{\rightarrow\dfrac{dy}{dx}=sec^2 x}$

$\tt{\rightarrow\dfrac{dy}{dx}(tanx)=sec^2 x}$