Let f(x)= | arcsin(t) dt and g(x)= | arccos(t) dt. | "'(x) g'(x) dæ =? 4 - T (a) 2 - T (b) 2 4 - T (c) (d) (e) -

Q: Find f. f'(t) 8 cos(t) + sec?(t), -7/2 <t < a/2, f(a/3) = 5 f(t) =

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Q: 3 Find f'(x)given f(x) = x/x – 4x – 7x %3D

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Q: 26 sin(x) dx. (Round your answers to six decimal places.) Jo (a) Find the approximations T. 10 M10…

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Q: JO (5) Find the derivatives (a) d # t ( * cos(2³) dz + 35e) 4

A: Since you have posted multiple questions i can do first question as per our company guidelines.…

Q: Find an equation of the line tangent to the function of x2 + (y – x)² = 9 at x 1.

A: You have posted multiple question. As per the instruction, we are answering the first one. If you…

Q: Calculate the Derivative k(u) = arc sec(u) 3u + 2

A: The given function is .The objective is to find the derivative of the given function.Derivative…

Q: Sketch the graph of f(x) = sin¬'(x² + y²) by analyzing its level curves.

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Q: Use the rule for square roots: f'(x) 2/ F(x) V F(x) to find the following derivatives (quickly d V…

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Q: Compute Each Derivative: 1) d tan¯¹ (1 + e²x) dx 2) x²cos-¹(3x4) d dx 3) x(sin-¹(x))² dx

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Q: a.) f(x) = sec(e") b.) g(x) = tan¯'(x)+3* +x!5 %3D _1.5

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Q: y=u (1+u³)* and u = (3x - 2)²

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Q: 3. Find the derivative of y with respect to x in the following functions : 18x'y (ans.: (3x-2) a)…

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Q: x(t)=sin2((pi/2)-t)  y(t)=cos4((pi/2-t) 0≤t≤pi/2 determine the slope of the curve at t= pi/4…

A: xt=sin2π2-t,  yt=cos4π2-t,  0≤t≤π2 Differentiating with respect to t, we get dxdt=2sinπ2-t·ddtπ2-t…

Q: find the second derivative of y=[(arctan((2+x)/(1-2x)))]

A: find the second derivative of y=[(arctan((2+x)/(1-2x)))]

Q: Find f. f(t) = f'(t) sec(t) (sec(t) + tan(t)), -2-√2 x Need Help? Read It Watch It - 1/2 < t < 1/₁ (…

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Q: Find f. f'(t) = sec(t)(sec(t) + tan(t)), f(t) = - 1/2 <²<= // ¹ ( ² ) = -3

A: Given that, f't=sectsect+tant, -π2&lt;t&lt;π2, fπ4=-3 We have to find the function ft

Q: Let f (t) = et sin 7t Part (a) Find L{f (t)} Part (b) Find L{f' (t)}

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Q: Find f. f'(t) = sec(t)(sec(t) + tan(t)), = -1 f(t) =

A: Given that: f'(t)=sect(sect+tant),    -π2&lt;t&lt;π2, fπ4=-1

Q: Q3) for f(t) = sec-(6t- 1) and g(t) = tan (6t - 1) show that s'() = g'(t). %3D

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