# Thread: derivative

1. ## derivative

Please help me to solve this
If $\displaystyle sin^2 A+sin^2 B+sin^2 C=K$ then obtain $\displaystyle \frac{dB} {dC}$
Thanks in advance

2. ## Re: derivative

let $\lambda = K-\sin^2(A)$

assume $\dfrac{dA}{dC} = 0$

$\sin^2(C) = \lambda - \sin^2(B)$

$\dfrac{d}{dC} \sin^2(C) = \dfrac{d}{dC} (\lambda - \sin^2(B))$

$2 \sin(C)\cos(C) = -2\sin(B)\cos(B)\dfrac{dB}{dC}$

$\dfrac{dB}{dC} =- \dfrac{ \sin(C)\cos(C)}{\sin(B)\cos(B)}=-\dfrac{\sin(2C)}{\sin(2B)}$

3. ## Re: derivative

Use implicit differentiation. Differentiate both sides with respect to C.

4. ## Re: derivative

Originally Posted by romsek
let $\lambda = K-\sin^2(A)$

assume $\dfrac{dA}{dC} = 0$

$\sin^2(C) = \lambda - \sin^2(B)$

$\dfrac{d}{dC} \sin^2(C) = \dfrac{d}{dC} (\lambda - \sin^2(B))$

$2 \sin(C)\cos(C) = -2\sin(B)\cos(B)\dfrac{dB}{dC}$

$\dfrac{dB}{dC} =- \dfrac{ \sin(C)\cos(C)}{\sin(B)\cos(B)}=-\dfrac{\sin(2C)}{\sin(2B)}$
Hello,
Assumption of $\frac{dA}{dC}$ is necessary.Without that we cann't compute $\frac{dB}{dC}$

5. ## Re: derivative

Originally Posted by Vinod
Hello,
Assumption of $\frac{dA}{dC}$ is necessary.Without that we cann't compute $\frac{dB}{dC}$
That is not true.

$$\sin 2A\cdot \dfrac{dA}{dC}+\sin 2B\cdot \dfrac{dB}{dC}+\sin 2C=\dfrac{dK}{dC}$$

$$\dfrac{dB}{dC} = \dfrac{\dfrac{dK}{dC}-\sin 2A\cdot \dfrac{dA}{dC}-\sin 2C}{\sin 2B}$$