Calculus I - Derivatives of common functions
21 flashcards · Mathematics · by luckydog789 (harrison holbrook)
ddxex\frac{d}{dx}e^xdxdex
ddxex=ex\frac{d}{dx}e^x=e^xdxdex=ex
ddxsin(x)\frac{d}{dx}\sin(x)dxdsin(x)
ddxsin(x)=cos(x)\frac{d}{dx}\sin(x)=\cos(x)dxdsin(x)=cos(x)
ddxcos(x)\frac{d}{dx}\cos(x)dxdcos(x)
ddxcos(x)=−sin(x)\frac{d}{dx}\cos(x)=-\sin(x)dxdcos(x)=−sin(x)
ddxtan(x)\frac{d}{dx}\tan(x)dxdtan(x)
ddxtan(x)=sec2(x)\frac{d}{dx}\tan(x)=\sec^2(x)dxdtan(x)=sec2(x)
ddxcot(x)\frac{d}{dx}\cot(x)dxdcot(x)
ddxcot(x)=−csc2(x)\frac{d}{dx}\cot(x)=-\csc^2(x)dxdcot(x)=−csc2(x)
ddxsec(x)\frac{d}{dx}\sec(x)dxdsec(x)
ddxsec(x)=sec(x)tan(x)\frac{d}{dx}\sec(x)=\sec(x)\tan(x)dxdsec(x)=sec(x)tan(x)
ddxcsc(x)\frac{d}{dx}\csc(x)dxdcsc(x)
ddxcsc(x)=−csc(x)cot(x)\frac{d}{dx}\csc(x)=-\csc(x)\cot(x)dxdcsc(x)=−csc(x)cot(x)
ddxln(x)\frac{d}{dx}\ln(x)dxdln(x)
ddxln(x)=1x\frac{d}{dx}\ln(x)=\frac{1}{x}dxdln(x)=x1 , for 0"> x>0x>0x>0 ddxln∣x∣=1x\frac{d}{dx}\ln|x|=\frac{1}{x}dxdln∣x∣=x1 , for x≠0x \neq 0x=0
ddxln∣u(x)∣\frac{d}{dx}\ln|u(x)|dxdln∣u(x)∣
ddxln∣u(x)∣=u′(x)u(x)\frac{d}{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}dxdln∣u(x)∣=u(x)u′(x)
ddxbx\frac{d}{dx}b^xdxdbx , for 0"> x>0x>0x>0
ddxbx=bxln(b)\frac{d}{dx}b^x=b^{x}\ln(b)dxdbx=bxln(b)
ddxg(x)h(x)\frac{d}{dx}g(x)^{h(x)}dxdg(x)h(x)
ddxg(x)h(x)=ddxeh(x)ln(g(x))\frac{d}{dx}g(x)^{h(x)}=\frac{d}{dx}e^{h(x)\ln(g(x))}dxdg(x)h(x)=dxdeh(x)ln(g(x))
ddxlogbx\frac{d}{dx}\log_{b}{x}dxdlogbx
ddxlogbx=1xln(b)\frac{d}{dx}\log_{b}{x}=\frac{1}{x\ln(b)}dxdlogbx=xln(b)1 , for 0"> x>0x>0x>0 ddxlogb∣x∣=1xln(b)\frac{d}{dx}\log_{b}{|x|}=\frac{1}{x\ln(b)}dxdlogb∣x∣=xln(b)1 , for x≠0x \neq 0x=0
ddxsin−1(x)\frac{d}{dx}\sin^{-1}(x)dxdsin−1(x)
ddxsin−1(x)=11−x2\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}dxdsin−1(x)=1−x2 1 ,for −1<x<1-1<x<1−1<x<1
ddxcos−1(x)\frac{d}{dx}\cos^{-1}(x)dxdcos−1(x)
ddxcos−1(x)=−11−x2\frac{d}{dx}\cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}dxdcos−1(x)=−1−x2 1 ,for −1<x<1-1<x<1−1<x<1
ddxtan−1(x)\frac{d}{dx}\tan^{-1}(x)dxdtan−1(x)
ddxtan−1(x)=11+x2\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}dxdtan−1(x)=1+x21 , for x∈Rx \in \mathbb{R}x∈R
ddxcot−1(x)\frac{d}{dx}\cot^{-1}(x)dxdcot−1(x)
ddxcot−1(x)=−11+x2\frac{d}{dx}\cot^{-1}(x)=-\frac{1}{1+x^2}dxdcot−1(x)=−1+x21 , for x∈Rx \in \mathbb{R}x∈R
ddxsec−1(x)\frac{d}{dx}\sec^{-1}(x)dxdsec−1(x)
ddxsec−1(x)=1∣x∣x2−1\frac{d}{dx}\sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}dxdsec−1(x)=∣x∣x2−1 1 , for 1"> ∣x∣>1|x|>1∣x∣>1
ddxcsc−1(x)\frac{d}{dx}\csc^{-1}(x)dxdcsc−1(x)
ddxcsc−1(x)=−1∣x∣x2−1\frac{d}{dx}\csc^{-1}(x)=-\frac{1}{|x|\sqrt{x^2-1}}dxdcsc−1(x)=−∣x∣x2−1 1 , for 1"> ∣x∣>1|x|>1∣x∣>1
ddxf−1(x)\frac{d}{dx}f^{-1}(x)dxdf−1(x)
ddxf−1(x)=1f′(x)\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(x)}dxdf−1(x)=f′(x)1 , for f′(x)≠0f'(x) \neq 0f′(x)=0
h(x)=f−1(x)h(x)=f^{-1}(x)h(x)=f−1(x) ddxh(x)\frac{d}{dx}h(x)dxdh(x)
ddxh(x)=1f′(h(x))\frac{d}{dx}h(x)=\frac{1}{f'(h(x))}dxdh(x)=f′(h(x))1 , for f′(h(x))≠0f'(h(x)) \neq 0f′(h(x))=0
ddx1g(x)\frac{d}{dx} \frac{1}{g(x)}dxdg(x)1
ddx1g(x)=−g′(x)g(x)2\frac{d}{dx} \frac{1}{g(x)}=\frac{-g'(x)}{g(x)^2}dxdg(x)1=g(x)2−g′(x)