Problem to solve
Derivate \[e^{\tan[x]}\]
Solution
STEP 1
Possible derivation: d/dx(exp(tan(x)))
STEP 2
Using the chain rule, d/dx(e^tan(x)) = ( d e^u)/( du) ( du)/( dx), where u = tan(x) and d/( du)(e^u) = e^u: = e^tan(x) (d/dx(tan(x)))
STEP 3
Using the chain rule, d/dx(tan(x)) = ( dtan(u))/( du) ( du)/( dx), where u = x and d/( du)(tan(u)) = sec^2(u): = (d/dx(x)) sec^2(x) e^tan(x)
STEP 4
The derivative of x is 1: = 1 e^tan(x) sec^2(x)
STEP 5
Simplify the expression: Answer: | | = e^tan(x) sec^2(x)
More problems
Number
001Derivate \(3\cos(x)-\frac{1}{\sqrt{x^3}}\)
002Derivate \(x\sqrt{x^5}+6\csc(x)+x^5\)
003Derivate \(2\cdot3^x+5\sqrt{x}-4\arccos(x)\)
004Derivate \(5\textrm{arccosh}(x)-4\cot(x)+2x^4\)
005Derivate \(e^{\tan(x)}\)
006Derivate \(\frac{\sec(x)-3x^4}{x^4+6}\)
007Derivate \(\frac{x^5+\tan(x)}{5^x-4\cos(x)}\)
008Derivate \(\frac{\ln(x)-\sec(x^4)}{6\cos(3x-6)-x^3}\)
009Derivate \(\tan(\cos^2(x))\)
010Derivate \(\sec^5(x^4-6x+9)\)
011Derivate \((\csc(x^4)-e^{x^5+6})(\tan(x^5)-5x^{4/5})\)
012Derivate \(\sec^3(x^2-cos(x^2))\)
013Derivate \(\sqrt{2x^3 \sec(x^4)+5}\)
014Derivate \(\arctan(x^2)-x^5\sin(x)\cos(x^5)\)
015Derivate \((x^4-6)^4(3x^7-6x^6+2)^5\)
016Derivate \(\frac{e^{6^x}}{\sqrt[4]{\sec^3(x+5)}}\)
017Derivate \(\csc(\tan^2(-3x^5-2))\)
018Derivate \(\textrm{arcsec}(x^4)\cdot7^{\sin(x)}\)
019Derivate \(\frac{\tanh(x^4)}{\sin^2(x)+1}\)
020Derivate \(\frac{(x^5-6x)^4}{(x^4+\cos(6x))^6}\)
021Derivate \(\ln(x^7-2x)\cdot\tan(x^4-2)\)
022Derivate \(\frac{(2x^5-2)^3(3x^6-1)^8}{(x^4-2)^5\sqrt{x^2-3}}\)
023Derivate \(\frac{\sin(\cos(x^3))}{\sqrt{\ln(x)}}\)
024Derivate \(\ln^4(x^5+\sin(x))\cdot\sin(4x)\)
025Derivate \(\tan(x^5)-\frac{7}{\sqrt{x^5+8}}\)
Back
Edited by tododemate® - 2021