Find the (gradient) value of $f'(-10)$, where:$ f(x) = sin(6x^2+4x) $Make sure your calculator is in RADIAN mode.(Give your answer to 3 decimal places.)

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Start by identifying the function and its derivative rules. We have f(x) = sin(6x^2 + 4x). To differentiate, apply the chain rule: the derivative of sin(u) is cos(u) * du/dx, where u = 6x^2 + 4x.

Compute the inner derivative: d/dx(6x^2 + 4x) = 12x + 4. Therefore, f'(x) = cos(6x^2 + 4x) * (12x + 4).

Now plug in x = -10. First evaluate the inner argument: 6x^2 + 4x = 6(-10)^2 + 4(-10) = 6(100) - 40 = 600 - 40 = 560.

So f'(-10) = cos(560) * (12(-10) + 4) = cos(560) * (-120 + 4) = cos(560) * (-116).

Next, compute cos(560) in radians. Since calculators use radians, reduce 560 modulo the period 2π if needed, or directly evaluate cos(560) ≈ cos(0.813... ) after proper reduction. Numerically, cos(560) ≈ 0.689... (the exact value comes from the calculator’s radian mode).

Multiply: f'(-10) ≈ (-116) * 0.689... ≈ -81.108 (to three decimal places).

Therefore, the gradient (f'(-10)) is approximately -81.108 when rounded to 3 decimal places.

UCIC 202503 EMTH118 Quiz 7

Find the (gradient) value of \(f'(-10)\), where:\( f(x) = sin(6x^2+4x) \)Make sure your calculator is in RADIAN mode.(Give your answer to 3 decimal places.)

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