3 ms·
You differentiate both sides of the equation given above with respect to t. Differentiating 1/t yields -1/t^2. On the LHS you apply the differentiation inside t
by mvanaltvorst 6y ago
You differentiate both sides of the equation given above with respect to t. Differentiating 1/t yields -1/t^2. On the LHS you apply the differentiation inside the integral.