Question

Determine a derivada de

a) f(x) = √x . cotgx . lnx
b)f(x) = 7^x √x / tgx

Answer 1

a) y = √x.(cotgx.lnx)

Aplique a regra do produto

y' = (√x).(cotgx.lnx)' + (√x)'.(cotgx.lnx)

Aplique novamente a mesma regra

(cotgx.lnx)' = (cotgx).(lnx)' + (cotgx)'.(lnx)

Sabemos que:

(ln x)' = 1/x

(cotgx)' = -cossec²x

(√x)' = 1/(2.√x)

Assim, teremos:

(cotgx.lnx)' = (cotgx).1/x + (-cossec²x).(lnx)

                  = (cotgx)/x - (cossec²x).(lnx)            

Logo,

y' = (√x).(cotgx.lnx)' + (√x)'.(cotgx.lnx)

y' = (√x).(cotgx/x - (cossec²x.lnx)) + 1/(2.√x).(cotgx.lnx)

y' = (√x).(cotgx - x(cossec²x.lnx) / x) + (cotgx.lnx)/(2.√x)    

Reorganizando:

$\frac{\cot \left(x\right)\ln \left(x\right)}{2\sqrt{x}}+\sqrt{x}\left(-\csc ^2\left(x\right)\ln \left(x\right)+\frac{\cot \left(x\right)}{x}\right)$

b) y =   (7^x .√x) / tgx

Regra do produto no numerador

(7^x.√x)' = (7^x)'.(√x) + (7^x).(√x)'

             = 7^x.ln 7.(√x) + (7^x).1/2(√x)

             = 7^x . ln 7 . √x + 7^x/2.√x

             = 7^x . (ln 7.√x + 1/(2√x))

Regra da divisão na função f(x) = y

y' = (7^x.√x)'.(tgx) - (tgx)'.(7^x.√x) / tg^2x

y' = {[7^x. tgx . (ln 7.√x + 1/(2√x))] - (-cossec²x).(7^x.√x)} /tg²x

y' = [[7^x . tgx . (ln 7.√x + 1/(2√x))] + cossec²x).(7^x.√x)} / tg²x

y' = [7^x. (tgx .(ln 7.√x + 1/(2√x)) + (√x).cossec²x) / tg²x

Reorganizando:

$\frac{\tan \left(x\right)\left(7^x+2\ln \left(7\right)\cdot \:7^xx\right)-2\cdot \:7^xx\sec ^2\left(x\right)}{2x^{\frac{1}{2}}\tan ^2\left(x\right)}}$