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- Step 4 Finally, determine 1 u du 1 dv + dx V dx d -(-3x + 7) and dx = 1 -3x + 7 = d dx 1 - (-3+²+7)( -3x + 7 d -(x + 7) and simplify the result. dx -(-3x + 7) + + 1 x + 7 1 (+ + 7) = x=( G d dx + -(x + 7) 1/17) (1) x + Therefore, if h(x) = In[(−3x + 7)(x + 7)], then we have the following result. d h'(x) = [In[(-3x + 7)(x + 7)]] = dxDifferentiate the following function. 6x2 +5 y = x2 +1 dx (x² + 1)< dy (x² + 1) • : (6x² + 5) + (6x² + 5) · (6x2 + 5) · (x2 + 1) dx dx dx d d (x2 + 1) dy (6x? + 5) – (6x² + 5) • (6x² + 5)2 -(2 +1)·은 (6x2+5)-(6x2 +5)·음 (2+1) -(x?+1) -- dx dx dx dy d d. (x² + 1) dx dx d d (x? + 1) • (6x2 + 5) - (6x? + 5) · - (x²+1) dy (x? +1) 2 dx dy dx IIDifferentiate the following: