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The Derivative Function Of The Reciprocal Of A Function

Let $s\left(x\right)=\ln x$ . What is

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{3x}$

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{2x}$

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{x}$

If $m\left(x\right)=\sin x$ , find .

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin^2x}$

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{-\cos x}{\sin^2x}$

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin x}$

Let . What is

$\left(\frac{1}{u}\right)'\left(x\right)=e^x$

$\left(\frac{1}{u}\right)'\left(x\right)=e^{-x}$

$\left(\frac{1}{u}\right)'\left(x\right)=2e^x$

Let $g\left(x\right)\ =\ \sqrt{x}$ . What is $\left(\frac{1}{g}\right)'\ \left(x\right)$ ?

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ \frac{1}{2\sqrt{x}}$

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -\frac{1}{2\sqrt{x}}$

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -2\sqrt{x}$

Let . What is

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos^2x}$

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos x}$

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{2\sin x}{\cos x}$

Consider $w\left(x\right)=\frac{1}{x^2}$ , determine $\left(\frac{1}{w}\right)'\left(x\right)$ .

$\left(\frac{1}{w}\right)'\left(x\right)=-\frac{2}{x^3}$

$\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x^3}$

$\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x}$

Let $q\left(x\right)=\frac{1}{x}$ . Determine ?

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^2}$

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^{ }}$

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{2x^2}$

If , determine

$\left(\frac{1}{v}\right)'\left(x\right)=-\frac{\cos^2x}{\tan^2x}$

$\left(\frac{1}{v}\right)'\left(x\right)=-2\frac{\cos^2x}{\tan^2x}$

$\left(\frac{1}{v}\right)'\left(x\right)=-3\frac{\cos^2x}{\tan^2x}$

Consider $h\left(x\right)=e^x$ . What is ?

$\left(\frac{1}{h}\right)'\left(x\right)=e^{-x}$

$\left(\frac{1}{h}\right)'\left(x\right)=e^x$

$\left(\frac{1}{h}\right)'\left(x\right)=e^{ }$

Given , determine

$\left(\frac{1}{r}\right)'\left(x\right)=-\frac{3}{x^4}$

$\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^4}$

$\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^{ }}$

Questions

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The Derivative Function Of The Reciprocal Of A Function

Let ﻿s(x)=ln⁡xs\left(x\right)=\ln xs(x)=lnx﻿ . What is

﻿(1s)′(x)=13x\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{3x}(s1​)′(x)=3x1​﻿

﻿(1s)′(x)=12x\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{2x}(s1​)′(x)=2x1​﻿

﻿(1s)′(x)=1x\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{x}(s1​)′(x)=x1​﻿

If ﻿m(x)=sin⁡xm\left(x\right)=\sin xm(x)=sinx﻿ , find .

﻿(1m)′(x)=cos⁡xsin⁡2x\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin^2x}(m1​)′(x)=sin2xcosx​﻿

﻿(1m)′(x)=−cos⁡xsin⁡2x\left(\frac{1}{m}\right)'\left(x\right)=\frac{-\cos x}{\sin^2x}(m1​)′(x)=sin2x−cosx​﻿

﻿(1m)′(x)=cos⁡xsin⁡x\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin x}(m1​)′(x)=sinxcosx​﻿

Let . What is

﻿(1u)′(x)=ex\left(\frac{1}{u}\right)'\left(x\right)=e^x(u1​)′(x)=ex﻿

﻿(1u)′(x)=e−x\left(\frac{1}{u}\right)'\left(x\right)=e^{-x}(u1​)′(x)=e−x﻿

﻿(1u)′(x)=2ex\left(\frac{1}{u}\right)'\left(x\right)=2e^x(u1​)′(x)=2ex﻿

Let ﻿g(x) = xg\left(x\right)\ =\ \sqrt{x}g(x) = x​﻿ . What is ﻿(1g)′ (x)\left(\frac{1}{g}\right)'\ \left(x\right)(g1​)′ (x)﻿?

﻿(1g)′ (x) = 12x\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ \frac{1}{2\sqrt{x}}(g1​)′ (x) = 2x​1​﻿

﻿(1g)′ (x) = −12x\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -\frac{1}{2\sqrt{x}}(g1​)′ (x) = −2x​1​﻿

﻿(1g)′ (x) = −2x\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -2\sqrt{x}(g1​)′ (x) = −2x​﻿

Let . What is

﻿(1t)′(x)=sin⁡xcos⁡2x\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos^2x}(t1​)′(x)=cos2xsinx​﻿

﻿(1t)′(x)=sin⁡xcos⁡x\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos x}(t1​)′(x)=cosxsinx​﻿

﻿(1t)′(x)=2sin⁡xcos⁡x\left(\frac{1}{t}\right)'\left(x\right)=\frac{2\sin x}{\cos x}(t1​)′(x)=cosx2sinx​﻿

Consider ﻿w(x)=1x2w\left(x\right)=\frac{1}{x^2}w(x)=x21​﻿ , determine ﻿(1w)′(x)\left(\frac{1}{w}\right)'\left(x\right)(w1​)′(x)﻿ .

﻿(1w)′(x)=−2x3\left(\frac{1}{w}\right)'\left(x\right)=-\frac{2}{x^3}(w1​)′(x)=−x32​﻿

﻿(1w)′(x)=2x3\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x^3}(w1​)′(x)=x32​﻿

﻿(1w)′(x)=2x\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x}(w1​)′(x)=x2​﻿

Let ﻿q(x)=1xq\left(x\right)=\frac{1}{x}q(x)=x1​﻿ . Determine ?

﻿(1q)′(x)=1x2\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^2}(q1​)′(x)=x21​﻿

﻿(1q)′(x)=1x\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^{ }}(q1​)′(x)=x1​﻿

﻿(1q)′(x)=12x2\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{2x^2}(q1​)′(x)=2x21​﻿

If , determine

﻿(1v)′(x)=−cos⁡2xtan⁡2x\left(\frac{1}{v}\right)'\left(x\right)=-\frac{\cos^2x}{\tan^2x}(v1​)′(x)=−tan2xcos2x​﻿

﻿(1v)′(x)=−2cos⁡2xtan⁡2x\left(\frac{1}{v}\right)'\left(x\right)=-2\frac{\cos^2x}{\tan^2x}(v1​)′(x)=−2tan2xcos2x​﻿

﻿(1v)′(x)=−3cos⁡2xtan⁡2x\left(\frac{1}{v}\right)'\left(x\right)=-3\frac{\cos^2x}{\tan^2x}(v1​)′(x)=−3tan2xcos2x​﻿

Consider ﻿h(x)=exh\left(x\right)=e^xh(x)=ex﻿ . What is ?

﻿(1h)′(x)=e−x\left(\frac{1}{h}\right)'\left(x\right)=e^{-x}(h1​)′(x)=e−x﻿

﻿(1h)′(x)=ex\left(\frac{1}{h}\right)'\left(x\right)=e^x(h1​)′(x)=ex﻿

﻿(1h)′(x)=e\left(\frac{1}{h}\right)'\left(x\right)=e^{ }(h1​)′(x)=e﻿

Given , determine

﻿(1r)′(x)=−3x4\left(\frac{1}{r}\right)'\left(x\right)=-\frac{3}{x^4}(r1​)′(x)=−x43​﻿

﻿(1r)′(x)=3x4\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^4}(r1​)′(x)=x43​﻿

﻿(1r)′(x)=3x\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^{ }}(r1​)′(x)=x3​﻿

Incorrect

Great!

Let $s\left(x\right)=\ln x$ . What is

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{3x}$

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{2x}$

$\left(\frac{1}{s}\right)'\left(x\right)=\frac{1}{x}$

If $m\left(x\right)=\sin x$ , find .

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin^2x}$

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{-\cos x}{\sin^2x}$

$\left(\frac{1}{m}\right)'\left(x\right)=\frac{\cos x}{\sin x}$

$\left(\frac{1}{u}\right)'\left(x\right)=e^x$

$\left(\frac{1}{u}\right)'\left(x\right)=e^{-x}$

$\left(\frac{1}{u}\right)'\left(x\right)=2e^x$

Let $g\left(x\right)\ =\ \sqrt{x}$ . What is $\left(\frac{1}{g}\right)'\ \left(x\right)$ ?

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ \frac{1}{2\sqrt{x}}$

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -\frac{1}{2\sqrt{x}}$

$\left(\frac{1}{g}\right)'\ \left(x\right)\ =\ -2\sqrt{x}$

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos^2x}$

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{\sin x}{\cos x}$

$\left(\frac{1}{t}\right)'\left(x\right)=\frac{2\sin x}{\cos x}$

Consider $w\left(x\right)=\frac{1}{x^2}$ , determine $\left(\frac{1}{w}\right)'\left(x\right)$ .

$\left(\frac{1}{w}\right)'\left(x\right)=-\frac{2}{x^3}$

$\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x^3}$

$\left(\frac{1}{w}\right)'\left(x\right)=\frac{2}{x}$

Let $q\left(x\right)=\frac{1}{x}$ . Determine ?

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^2}$

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{x^{ }}$

$\left(\frac{1}{q}\right)'\left(x\right)=\frac{1}{2x^2}$

$\left(\frac{1}{v}\right)'\left(x\right)=-\frac{\cos^2x}{\tan^2x}$

$\left(\frac{1}{v}\right)'\left(x\right)=-2\frac{\cos^2x}{\tan^2x}$

$\left(\frac{1}{v}\right)'\left(x\right)=-3\frac{\cos^2x}{\tan^2x}$

Consider $h\left(x\right)=e^x$ . What is ?

$\left(\frac{1}{h}\right)'\left(x\right)=e^{-x}$

$\left(\frac{1}{h}\right)'\left(x\right)=e^x$

$\left(\frac{1}{h}\right)'\left(x\right)=e^{ }$

$\left(\frac{1}{r}\right)'\left(x\right)=-\frac{3}{x^4}$

$\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^4}$

$\left(\frac{1}{r}\right)'\left(x\right)=\frac{3}{x^{ }}$