differentiate y = 1/4 (tan^4 x) Topic: Math 1 Answer | Add Yours jeew-m | College Teacher | (Level 1) Educator Emeritus Posted on July 1, 2013 at 5:18 AM `y = 1/4tan^4x` `(dy)/dx ` `= (d(1/4tan^4x))/dx` `= 1/4(d(tan^4x))/dx` `= 1/4xx4tan^3x(d(tanx))/dx` `= 1/4xx4tan^3xsec^2x` `= tan^3xsec^2x` So the answer is `(dy)/dx = tan^3xsec^2x` `y = 1/4tan^4x` `(dy)/dx ` `= (d(1/4tan^4x))/dx` `= 1/4(d(tan^4x))/dx` `= 1/4xx4tan^3x(d(tanx))/dx` `= 1/4xx4tan^3xsec^2x` `= tan^3xsec^2x` So the answer is `(dy)/dx = tan^3xsec^2x` We’ve answered 300,944 questions. We can answer yours, too. Ask a question

jeew-m | College Teacher | (Level 1) Educator Emeritus Posted on July 1, 2013 at 5:18 AM `y = 1/4tan^4x` `(dy)/dx ` `= (d(1/4tan^4x))/dx` `= 1/4(d(tan^4x))/dx` `= 1/4xx4tan^3x(d(tanx))/dx` `= 1/4xx4tan^3xsec^2x` `= tan^3xsec^2x` So the answer is `(dy)/dx = tan^3xsec^2x` `y = 1/4tan^4x` `(dy)/dx ` `= (d(1/4tan^4x))/dx` `= 1/4(d(tan^4x))/dx` `= 1/4xx4tan^3x(d(tanx))/dx` `= 1/4xx4tan^3xsec^2x` `= tan^3xsec^2x` So the answer is `(dy)/dx = tan^3xsec^2x`