Question

Differentiate the equation with respect to x:
cosx+sin2x.

Answer 1

In order to solve this problem differentiate the equation with respect to x. Be careful about signs. Doing this will solve your problem. Differentiation of cosx is - sinx and sinx is cosx.

Complete step-by-step answer:
The differentiation is nothing but the derivative is the instantaneous rate of function change with respect to one of its variables. This is equivalent to the discovery of the slope of the tangent line to the function at a point.

The given equation is :
$\Rightarrow$cosx+sin2x
We know that differentiation of cosx with respect to x is –sinx and that of sinx is cosx.
$\Rightarrow$ $\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx + sin2x)}}$ ($\dfrac{{\text{d}}}{{{\text{dx}}}}$(p) means differentiation of p with respect to x)
$\Rightarrow \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx) + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sin2x)}}$ $\left( \begin{gathered} \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sin2x)}} = 2\cos 2{\text{x, because we are differentiating the term with }} \\\ {\text{respect to x but there is a term 2x}}{\text{. As 2x is another function we will differentiate }} \\\ {\text{2x again then we get 2 since }}\dfrac{{\text{d}}}{{{\text{dx}}}}(2{\text{x) = 2}}{\text{.}} \\\ \end{gathered} \right)$
$\Rightarrow$-sinx+cos2x $\dfrac{{\text{d}}}{{{\text{dx}}}}(2{\text{x)}}$
$\Rightarrow - \sin {\text{x + 2cos2x}}$
$\Rightarrow$2cos2x-sinx is the answer to the given question.

Note: Whenever you face such types of problems you have to use the formulas of differentiation of trigonometric functions. As we know differentiation of cosx with respect to x is –sinx and that of sinx is cosx. Differentiation is opposite of integration. integrating the term which you have differentiated here with respect to the same variable you will get the same equation as given in question.