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Find f '(−4), if f(x) = (5x2 + 6x)(3x2 + 7). Round your answer to the nearest integer.

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Use:\\(h(x)\cdot g(x))'=h'(x)g(x)+h(x)g'(x)\\\\h(x)=5x^2+6x\to h'(x)=(5x^2+6x)'=10x+6\\\\g(x)=3x^2+7\to g'(x)=(3x^2+7)'=6x\\\\f(x)=(5x^2+6x)(3x^2+7)\\\\f'(x)=(5x^2+6x)'(3x^2+7)+(5x^2+6x)(3x^2+7)'\\\\=(10x+6)(3x^2+7)+(5x^2+6x)(6x)\\\\\text{use distributive property}\\\\=30x^3+70x+18x^2+42+30x^3+36x^2\\\\=60x^3+54x^2+70x+42


\text{Put -4 instead of x into the expression}\\\\f'(-4)=60(-4)^3+54(-4)^2+70(-4)+42\\\\=60(-64)+54(16)-280+42=-3214

User Johannes Flood
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