Q.
Let f(x)=∣∣sin3xcos3xtan3x1−142(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x∣∣
Then, the value of f′(x) at x=(2n+1)π,n∈I (the set of integers) is equal to
Given, f(x)=∣∣<br/>sin3x<br/>cos3x<br/>tan3x1−142(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x<br/>∣∣
On differentiating w.r.t. x, we get f′(x)=∣∣dxd(sin3x)dxd(cos3x)dxd(tan3x)1−142(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x∣∣ +∣∣sin3xcos3xtan3xdxd(1)dxd(−1)dxd(4)2(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x∣∣ +∣∣sin3xcos3xtan3x1−142dxd(cos23x+sin23x)22dxd(cos223x−sin223x)dxd(1+2tan3x)∣∣ =∣∣3cos3x3sin3x3sec23x1−142(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x∣∣ +∣∣sin3xcos3xtan3x0002(cos23x+sin23x)22(cos223x−sin223x)1+2tan3x∣∣ +∣∣sin3xcos3xtan3x1−142×2(cos23x+sin23x)×(2−3sin23x+23cos23x)2(−2cos23x×23sin23x−2sin23x×23cos23x)(0+2×3sec23x)∣∣
At x=(2n+1)π f′(x)=∣∣3(−1)031−142(1)2(−1)1+0∣∣+0 +∣∣0−101−144(0−1)×[−23(−1)+23×0]00∣∣ =[−3(−1+8)−1(0+6)+2(0+3)]+[0−1(0−0)−6(−4)] =−21+24=3