Question

find the derivative of y=sinx+tanx​

Answer 1

Problem decoded dude

answer is in attachment

thank meh :)

Answer 2

Given:

y= sinx+tanx

To Find:

Derivative of y with respect to x or dy/dx

Solution:

We know that,

$\longrightarrow\rm{\dfrac{d}{dx}(f(x)\pm g(x))=\dfrac{d}{dx}(f(x))\pm \dfrac{d}{dx}(g(x))}$

$\longrightarrow\rm{\dfrac{d}{dx}(sinx)=cosx}$

$\longrightarrow\rm{\dfrac{d}{dx}(tanx)=sec^{2}x}$

$\rule{190}{1}$

On differentiating y with respect to x, we get

$\longrightarrow\rm\green{\dfrac{dy}{dx}=cosx+sec^{2}x}$

$\rule{190}{1}$

Formulae to Remember

$\longrightarrow\rm{\dfrac{d}{dx}(cosx)=-sinx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cotx)=-cosec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(secx)=secx.tanx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosecx)=-cosecx.cotx}$

$\longrightarrow\rm{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

$\longrightarrow\rm{\dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(e^{x})=e^{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(f(x))=f'(x)}$