1. ## differentiation help (again)

trying to find the derivative of sec-1 (-5^(x^2+1))

Sorry, if it appears like it's not right side up, just click on the photo twice

2. ## Re: differentiation help (again)

\displaystyle \begin{align*} y &= \textrm{arcsec}\,\left( -5^{x^2 + 1} \right) \\ \sec{ \left( y \right) } &= -5^{x^2 + 1} \\ -\sec{ \left( y \right) } &= \mathrm{e}^{\ln{ \left( 5^{x^2 + 1} \right) } } \\ -\sec{ \left( y \right) } &= \mathrm{e}^{ \left( x^2 + 1 \right) \ln{ \left( 5 \right) } } \\ \frac{\mathrm{d}}{\mathrm{d}x} \left[ -\sec{ \left( y \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \mathrm{e}^{ \left( x^2 + 1 \right) \ln{ \left( 5 \right) } } \right] \\ -\sec{ \left( y \right) } \tan{ \left( y \right) } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \, x \, \mathrm{e}^{\left( x^2 + 1 \right) \ln{ \left( 5 \right) }} \\ -\sec{ \left( y \right) } \,\sqrt{ \sec^2{ \left( y \right) } - 1 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x\,\mathrm{e}^{ \ln{ \left( 5^{x^2 + 1} \right) } } \\ - \left( -5^{x^2 + 1} \right) \,\sqrt{ \left( -5^{x^2 + 1} \right) ^2 - 1 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x \, 5^{x^2 + 1} \\ 5^{x^2 + 1}\,\sqrt{ 5^{ 2\,x^2 + 2 } - 1 }\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x\,5^{x^2 + 1} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2\ln{ \left( 5 \right) }\,x}{\sqrt{ 5^{2\,x^2 + 2} - 1 }} \end{align*}

3. ## Re: differentiation help (again)

Originally Posted by Prove It
\displaystyle \begin{align*} y &= \textrm{arcsec}\,\left( -5^{x^2 + 1} \right) \\ \sec{ \left( y \right) } &= -5^{x^2 + 1} \\ -\sec{ \left( y \right) } &= \mathrm{e}^{\ln{ \left( 5^{x^2 + 1} \right) } } \\ -\sec{ \left( y \right) } &= \mathrm{e}^{ \left( x^2 + 1 \right) \ln{ \left( 5 \right) } } \\ \frac{\mathrm{d}}{\mathrm{d}x} \left[ -\sec{ \left( y \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \mathrm{e}^{ \left( x^2 + 1 \right) \ln{ \left( 5 \right) } } \right] \\ -\sec{ \left( y \right) } \tan{ \left( y \right) } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \, x \, \mathrm{e}^{\left( x^2 + 1 \right) \ln{ \left( 5 \right) }} \\ -\sec{ \left( y \right) } \,\sqrt{ \sec^2{ \left( y \right) } - 1 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x\,\mathrm{e}^{ \ln{ \left( 5^{x^2 + 1} \right) } } \\ - \left( -5^{x^2 + 1} \right) \,\sqrt{ \left( -5^{x^2 + 1} \right) ^2 - 1 } \,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x \, 5^{x^2 + 1} \\ 5^{x^2 + 1}\,\sqrt{ 5^{ 2\,x^2 + 2 } - 1 }\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 2\ln{ \left( 5 \right) } \,x\,5^{x^2 + 1} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2\ln{ \left( 5 \right) }\,x}{\sqrt{ 5^{2\,x^2 + 2} - 1 }} \end{align*}
how did you go from tany to the sqrt(sec) part? Did you use the triangle to find tan(y) ? And how did you get from -5^(2x^2+2) to 5^(2x^2+2), where did the negative go? Is it because you squared it,so it would be positive anyway? Also, is this the only way to do by rewriting inverse?

Thanks, i wasn't sure ;/

4. ## Re: differentiation help (again)

Do you at least know that $sin^2(y)+ cos^2(y)= 1$? If so then divide both sides by $cos^2(y)$ to get $\frac{sin^2(y)}{cos^2(y)}+ \frac{cos^2(y)}{cos^2(y)}= \frac{1}{cos^2(y)}$ which is equivalent to $tan^2(y)+ 1= sec^2(y)$ so that $tan^2(y)= sec^2(y)- 1$ and then $tan(x)= \pm\sqrt{sec^2(y)- 1}$.

Yes, because he squared $\left(-5^{x^2+ 1}\right)^2$ the negative sign "disappears": $(-1)^2= 1$. Also $\left(5^{x^2+ 1}\right)^2= 5^{2(x^2+ 1)}= 5^{2x^2+ 2}$ because $(a^b)^c= a^{bc}$.

No, you don't have to "rewrite inverse". You could, instead, have used the fact that $\frac{d(arcsec(y)}{dx}= \frac{1}{|x|\sqrt{x^2- 1}}$ (using a "table of derivatives") but generally, using "implicit differentiation" is simpler for inverse trig functions or, for that matter, inverse functions in general.

5. ## Re: differentiation help (again)

Originally Posted by HallsofIvy
Do you at least know that $sin^2(y)+ cos^2(y)= 1$? If so then divide both sides by $cos^2(y)$ to get $\frac{sin^2(y)}{cos^2(y)}+ \frac{cos^2(y)}{cos^2(y)}= \frac{1}{cos^2(y)}$ which is equivalent to $tan^2(y)+ 1= sec^2(y)$ so that $tan^2(y)= sec^2(y)- 1$ and then $tan(x)= \pm\sqrt{sec^2(y)- 1}$.

Yes, because he squared $\left(-5^{x^2+ 1}\right)^2$ the negative sign "disappears": $(-1)^2= 1$. Also $\left(5^{x^2+ 1}\right)^2= 5^{2(x^2+ 1)}= 5^{2x^2+ 2}$ because $(a^b)^c= a^{bc}$.

No, you don't have to "rewrite inverse". You could, instead, have used the fact that $\frac{d(arcsec(y)}{dx}= \frac{1}{|x|\sqrt{x^2- 1}}$ (using a "table of derivatives") but generally, using "implicit differentiation" is simpler for inverse trig functions or, for that matter, inverse functions in general.
how do i know whether i should keep the + or - in sqrt(sec^2x-1)?