Solve the derivative of expression: tan^2(ln(square root(4e^x+2x^2))) + cot^2(ln(square root(4e^x+2x^2)))

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have the expression: tan( ln (sqrt (4e^x+2x^2)))^2 + cot (ln (sqrt (4e^x + 2x^2)))^2

We have to find the derivative of the expression. We use the chain rule and start from the innermost function.

f(x) = tan( ln (sqrt (4e^x+2x^2)))^2 + cot (ln (sqrt (4e^x + 2x^2)))^2

f'(x) = 2*tan( ln (sqrt (4e^x+2x^2)))* (sec(ln (sqrt (4e^x+2x^2))))^2*(1/sqrt (4e^x+2x^2))*(1/2)*(1/sqrt (4e^x+2x^2))*(4e^x + 4x) + 2*cot (ln (sqrt (4e^x + 2x^2)))*(cosec(ln (sqrt (4e^x+2x^2))))^2*(1/sqrt (4e^x+2x^2))*(1/2)*(1/sqrt (4e^x+2x^2))*(4e^x + 4x)

=> [2*tan( ln (sqrt (4e^x+2x^2)))* (sec(ln (sqrt (4e^x+2x^2))))^2* + 2*cot (ln (sqrt (4e^x + 2x^2)))*(cosec(ln (sqrt (4e^x+2x^2))))^2]*(1/sqrt (4e^x+2x^2))*(1/2)*(1/sqrt (4e^x+2x^2))*(4e^x + 4x)

=> [2*tan( ln (sqrt (4e^x+2x^2)))* (sec(ln (sqrt (4e^x+2x^2))))^2* + 2*cot (ln (sqrt (4e^x + 2x^2)))*(cosec(ln (sqrt (4e^x+2x^2))))^2]*(1/(4e^x+2x^2))*(1/2)*(4e^x + 4x)

The required derivative is [2*tan( ln (sqrt (4e^x+2x^2)))* (sec(ln (sqrt (4e^x+2x^2))))^2* + 2*cot (ln (sqrt (4e^x + 2x^2)))*(cosec(ln (sqrt (4e^x+2x^2))))^2]*(1/(4e^x+2x^2))*(1/2)*(4e^x + 4x)

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that each term of the sum is a composed function, so, we'll calculate the derivative of each term.

We'll note the first term as f(x) and the 2nd term as g(x).

f(x) = tan^2(ln(square root(4e^x+2x^2)))

To determine the derivative of f(x), we'll apply the chain rule:

f'(x) = 2tan{ln[sqrt(4e^x+2x^2)]}*(1/[cos(lnsqrt(4e^x+2x^2)]^2*[1/2sqrt(4e^x+2x^2)]*(4e^x + 4x)

We'll calculate the derivative of g(x):

g'(x) = -2cot{ln[sqrt(4e^x+2x^2)]}*(1/[sin(lnsqrt(4e^x+2x^2)]^2*[1/2sqrt(4e^x+2x^2)]*(4e^x + 4x)

Now, we'll add f'(x) + g'(x):

tan{ln[sqrt(4e^x+2x^2)]}*(1/[cos(lnsqrt(4e^x+2x^2)]^2*[1/sqrt(4e^x+2x^2)]*(4e^x + 4x) - cot{ln[sqrt(4e^x+2x^2)]}*(1/[sin(lnsqrt(4e^x+2x^2)]^2*[1/sqrt(4e^x+2x^2)]*(4e^x + 4x)

We'll factorize by (4e^x + 4x)/sqrt(4e^x+2x^2)

 f'(x) + g'(x) = [(4e^x + 4x)/sqrt(4e^x+2x^2)]*{tan ln[sqrt(4e^x+2x^2)]/[cos(lnsqrt(4e^x+2x^2)]^2] - cot ln[sqrt(4e^x+2x^2)]/[sin(lnsqrt(4e^x+2x^2)]^2}

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question