Given function
y = ∑ k = 1 6 k cos − 1 ( 3 5 cos k x − 4 5 sin k x ) y=\sum_{k=1}^{6} k\cos^{-1}\left(\frac35\cos kx-\frac45\sin kx\right) y = k = 1 ∑ 6 k cos − 1 ( 5 3 cos k x − 5 4 sin k x ) We need to find d y d x \dfrac{dy}{dx} d x d y x = 0 x=0 x = 0
Differentiate a general term
Let
T k = k cos − 1 ( 3 5 cos k x − 4 5 sin k x ) T_k = k\cos^{-1}\left(\frac35\cos kx-\frac45\sin kx\right) T k = k cos − 1 ( 5 3 cos k x − 5 4 sin k x ) Set
u k = 3 5 cos k x − 4 5 sin k x u_k=\frac35\cos kx-\frac45\sin kx u k = 5 3 cos k x − 5 4 sin k x Then
d T k d x = k ( − 1 1 − u k 2 ) d u k d x \frac{dT_k}{dx}=k\left(-\frac{1}{\sqrt{1-u_k^2}}\right)\frac{du_k}{dx} d x d T k = k ( − 1 − u k 2 1 ) d x d u k Now,
d u k d x = 3 5 ( − k sin k x ) − 4 5 ( k cos k x ) = − 3 k 5 sin k x − 4 k 5 cos k x \frac{du_k}{dx}=\frac35(-k\sin kx)-\frac45(k\cos kx)
=-\frac{3k}{5}\sin kx-\frac{4k}{5}\cos kx d x d u k = 5 3 ( − k sin k x ) − 5 4 ( k cos k x ) = − 5 3 k sin k x − 5 4 k cos k x So,
d T k d x = k ⋅ ( − 1 1 − u k 2 ) ( − 3 k 5 sin k x − 4 k 5 cos k x ) \frac{dT_k}{dx}=k\cdot \left(-\frac{1}{\sqrt{1-u_k^2}}\right)\left(-\frac{3k}{5}\sin kx-\frac{4k}{5}\cos kx\right) d x d T k = k ⋅ ( − 1 − u k 2 1 ) ( − 5 3 k sin k x − 5 4 k cos k x ) Hence,
d T k d x = k 2 ( 3 5 sin k x + 4 5 cos k x ) 1 − ( 3 5 cos k x − 4 5 sin k x ) 2 \frac{dT_k}{dx}=\frac{k^2\left(\frac35\sin kx+\frac45\cos kx\right)}{\sqrt{1-\left(\frac35\cos kx-\frac45\sin kx\right)^2}} d x d T k = 1 − ( 5 3 cos k x − 5 4 sin k x ) 2 k 2 ( 5 3 sin k x + 5 4 cos k x )
Evaluate at x = 0 x=0 x = 0
At x = 0 x=0 x = 0
cos 0 = 1 , sin 0 = 0 \cos 0=1,\qquad \sin 0=0 cos 0 = 1 , sin 0 = 0 Therefore,
u k ( 0 ) = 3 5 u_k(0)=\frac35 u k ( 0 ) = 5 3 and
1 − u k ( 0 ) 2 = 1 − ( 3 5 ) 2 = 1 − 9 25 = 16 25 = 4 5 \sqrt{1-u_k(0)^2}=\sqrt{1-\left(\frac35\right)^2}
=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac45 1 − u k ( 0 ) 2 = 1 − ( 5 3 ) 2 = 1 − 25 9 = 25 16 = 5 4 Also,
3 5 sin 0 + 4 5 cos 0 = 4 5 \frac35\sin 0+\frac45\cos 0=\frac45 5 3 sin 0 + 5 4 cos 0 = 5 4 Thus,
\left.\frac{dT_k}{dx}\right|_{x=0}= rac{k^2\cdot \frac45}{\frac45}=k^2
Sum over k = 1 k=1 k = 1 6 6 6
So,
d y d x ∣ x = 0 = ∑ k = 1 6 k 2 \left.\frac{dy}{dx}\right|_{x=0}=\sum_{k=1}^{6} k^2 d x d y x = 0 = k = 1 ∑ 6 k 2 Now,
1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 = 1 + 4 + 9 + 16 + 25 + 36 = 91 1^2+2^2+3^2+4^2+5^2+6^2=1+4+9+16+25+36=91 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
Final answer
91 \boxed{91} 91 The derived answer matches the stored correct answer.