What is (d^2)/(dx^2)(5cos(2x+10)) ?

The answer to (d^2)/(dx^2)(5cos(2x+10)) is -20cos(2x+10)

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(d^2)/(dx^2)(5cos(2x+10))

\frac{d ^{2}}{d x ^{2}} (5 cos (2 x + 10))

- 20 cos (2 x + 10)

Solution steps

\frac{d ^{2}}{d x ^{2}} (5 cos (2 x + 10))

\frac{d}{d x} (5 cos (2 x + 10)) = - 10 sin (2 x + 10)

= \frac{d}{d x} (- 10 sin (2 x + 10))

\frac{d}{d x} (- 10 sin (2 x + 10)) = - 20 cos (2 x + 10)

= - 20 cos (2 x + 10)

q \to 18 lim (0.9 q^{2} - 6 q + 80)

sec (θ) (θ - tan (θ))

x^{2} + 8

f (x) = x^{2} - 10 x + 34 + x^{2} + 16

\int (x - \frac{3}{x})^{2} x d x

What is (d^2)/(dx^2)(5cos(2x+10)) ?
The answer to (d^2)/(dx^2)(5cos(2x+10)) is -20cos(2x+10)