Question

Maths formula for differential calculus

Answer 1

CALCULUS FORMULAS

Calculus is one of the branches of Mathematics that involves in the study of ‘Rage to Change’ and their application to solving equations. It has two major branches, Differential Calculus that is concerning rates of change and slopes of curves, and Integral Calculus concerning accumulation of quantities and the areas under and between curves.

Both branches make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. These two branches are related to each other by the fundamental theorem of calculus

The Differential Calculus splits up an area into small parts to calculate the rate of change. While, the Integral calculus joins small parts to calculates the area or volume. In short, it is the method of reasoning or calculation.

In this page you can see a list of Calculus Formulas such as integral formula, derivative formula, limits formula etc.

ddxrn=nxn−1ddxrn=nxn−1

ddx(fg)=fg1+gf1ddx(fg)=fg1+gf1

ddx(fg)=gf1−fg1g2ddx(fg)=gf1−fg1g2

ddxf(g(x))=f1(g(x))g1(x)ddxf(g(x))=f1(g(x))g1(x)

ddx(sinx)=cosxddx(sinx)=cosx

ddx(cosx)=−sinxddx(cosx)=−sinx

ddx(tanx)=−sec2xddx(tanx)=−sec2x

ddx(cotx)=csc2xddx(cotx)=csc2x

ddx(secx)=secxtanxddx(secx)=secxtanx

ddx(cscx)=−cscxcotxddx(cscx)=−cscxcotx

ddx(ex)=exddx(ex)=ex

ddx(ax)=axlnaddx(ax)=axlna

ddxlnx=1xddxlnx=1x

ddx(arcsinx)=11−x2−−−−−√ddx(arcsin⁡x)=11−x2

ddx(arcsinx)=11+x2ddx(arcsin⁡x)=11+x2

Integration Formulas

∫adr=ax+C∫adr=ax+C

∫1xdr=ln|x|+C∫1xdr=ln⁡|x|+C

∫exdx=ex+C∫exdx=ex+C

∫axdx=exlna+C∫axdx=exln⁡a+C

∫lnxdx=xlnx−x+C∫ln⁡xdx=xln⁡x−x+C

∫sinxdx=−cosx+C∫sinxdx=−cosx+C

∫cosxdx=sinx+C∫cosxdx=sinx+C

∫tandr+ln|secx|+Cor−ln|cosx|+C∫tandr+ln⁡|secx|+Cor−ln⁡|cosx|+C

∫cotxdr=ln|sinx|+C∫cotxdr=ln⁡|sinx|+C

∫secxdx=ln|secx+tanx|+C∫secxdx=ln⁡|secx+tanx|+C

∫cscxdx=ln|cscx–cotx|+C∫cscxdx=ln⁡|cscx–cotx|+C

∫sec2xdx=tanx+C∫sec2xdx=tanx+C

∫secxtanxdx=secx+C∫secxtanxdx=secx+C

∫csc2xdr=−cotx+C∫csc2xdr=−cotx+C

∫tan2xdr=tanx−x+C∫tan2xdr=tanx−x+C

∫dra2−x2−−−−−−√=arcsin(xa)+C∫dra2−x2=arcsin⁡(xa)+C

∫dra2+x2−−−−−−√=1aarcsin(xa)+C

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