identify-the-inner-function-ugxu-gxugx-and-the-outer-function-yfuy-fu-yfu-for-th

Question

Identify the inner function ﻿u=g(x)u = g(x)u=g(x)﻿, and the outer function ﻿y=f(u)y = f(u) y=f(u)﻿ for the composition ﻿y=f(g(x))y = f(g(x))y=f(g(x))﻿. Then, find ﻿dydx\frac{\mathrm{dy}}{\mathrm{dx}}dxdy​﻿:y=tan⁡(8x4)y=	an\$$left(8x^4$$ight)

Accepted Answer

u=8x4u=$$8x^4$$; f(u)=tan⁡uf\left(uight)=	an u; dydx=32x3sec⁡2(8x4)\frac{\mathrm{dy}}{\mathrm{dx}}=$$32x^3$$\$$sec^2$$\$$left(8x^4$$ight)

Answer

u=8x4u=$$8x^4$$; f(u)=sec⁡uf\left(u\right)=\sec u; dydx=32x3sec⁡2(8x4)\frac{\mathrm{dy}}{\mathrm{dx}}=$$32x^3$$\$$sec^2$$\$$left(8x^4$$\right)

Answer

u=8xu=8x; f(u)=sec⁡uf\left(u\right)=\sec u; dydx=32x3sec⁡2(8x4)\frac{\mathrm{dy}}{\mathrm{dx}}=$$32x^3$$\$$sec^2$$\$$left(8x^4$$\right)

Answer

u=8x4u=$$8x^4$$; f(u)=tan⁡uf\left(uight)=	an u; dydx=32x4sec⁡2(8x4)\frac{\mathrm{dy}}{\mathrm{dx}}=$$32x^4$$\$$sec^2$$\$$left(8x^4$$ight)

Identify the inner function u=g(x)u = g(x)u=g(x), and the outer function y=f(u)y = f(u) y=f(u) for the composition y=f(g(x))y = f(g(x))y=f(g(x)). Then, find dydx\(\frac{\mathrm{dy}\)}{\(\mathrm{dx}\)}dxdy:y=tan(8x4)y=\(\tan\]\left\)(8x^4\(\right\))

Identify the inner function $u = g(x)$ , and the outer function $y = f(u)$ for the composition $y = f(g(x))$ . Then, find $\(\frac{\mathrm{dy}\)}{\(\mathrm{dx}\)}$ :
$y = \tan (8 x^{4})$