Find the derivatives of the functions g(t)=(\frac{1+\sin 3t}{3-2t})^{-1}

Cem Hayes

Cem Hayes

Answered question

2021-10-18

Find the derivatives of the functions
g(t)=(1+sin3t32t)1

Answer & Explanation

Bertha Stark

Bertha Stark

Skilled2021-10-19Added 96 answers

Given,
g(t)=(1+sin3t32t)1
g(t)=32t1+sin3t
Differentiating with respect to t, we get
g(t)=d(32t)dt(1+sin3t)d(1+sin3t)dt(32t)(1+sin3t)2
g(t)=2(1+sin3t)3cos3t(32t)(1+sin3t)2
g(t)=22sin3t9cos3t+6tcos3t(1+sin3t)2

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