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Exercise 8.4 · Q1

Q.Find the partial derivatives of the following functions at the indicated points.

(i) f(x,y)=3x2−2xy+y2+5x+2,(2,−5)f(x,y)=3x^2-2xy+y^2+5x+2,\quad (2,-5)
(ii) g(x,y)=3x2+y2+5x+2,(1,−2)g(x,y)=3x^2+y^2+5x+2,\quad (1,-2)
(iii) h(x,y,z)=xsin⁡(xy)+z2x,(2,π4,1)h(x,y,z)=x\sin(xy)+z^2x,\quad \left(2,\dfrac{\pi}{4},1\right)
(iv) G(x,y)=ex+3ylog⁡(x2+y2),(−1,1)G(x,y)=e^{x+3y}\log(x^2+y^2),\quad(-1,1)
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✓ Free question

For each function, hold every variable except the one being differentiated constant, apply the ordinary derivative rules, then substitute the given point.

Part (i): f(x,y)=3x2−2xy+y2+5x+2f(x,y)=3x^2-2xy+y^2+5x+2 at (2,−5)(2,-5).

fx=6x−2y+5f_x=6x-2y+5. At (2,−5)(2,-5): 6(2)−2(−5)+5=12+10+5=276(2)-2(-5)+5=12+10+5=27.

fy=−2x+2yf_y=-2x+2y. At (2,−5)(2,-5): −2(2)+2(−5)=−4−10=−14-2(2)+2(-5)=-4-10=-14.

Part (ii): g(x,y)=3x2+y2+5x+2g(x,y)=3x^2+y^2+5x+2 at (1,−2)(1,-2).

gx=6x+5g_x=6x+5. At (1,−2)(1,-2): 6(1)+5=116(1)+5=11.

gy=2yg_y=2y. At (1,−2)(1,-2): 2(−2)=−42(-2)=-4.

Part (iii): h(x,y,z)=xsin⁡(xy)+z2xh(x,y,z)=x\sin(xy)+z^2x at (2,π4,1)\left(2,\dfrac\pi4,1\right).

hx=sin⁡(xy)+x⋅cos⁡(xy)⋅y+z2=sin⁡(xy)+xycos⁡(xy)+z2h_x=\sin(xy)+x\cdot\cos(xy)\cdot y+z^2 = \sin(xy)+xy\cos(xy)+z^2 (product rule on xsin⁡(xy)x\sin(xy), plus z2z^2 from z2xz^2x).

At the point: xy=2⋅π4=π2xy=2\cdot\dfrac\pi4=\dfrac\pi2, so sin⁡(xy)=1, cos⁡(xy)=0\sin(xy)=1,\ \cos(xy)=0. hx=1+(π2)(0)+12=1+0+1=2h_x=1+\left(\dfrac\pi2\right)(0)+1^2=1+0+1=2.

hy=x⋅cos⁡(xy)⋅x=x2cos⁡(xy)h_y=x\cdot\cos(xy)\cdot x=x^2\cos(xy) (only the xsin⁡(xy)x\sin(xy) term depends on yy). At the point: x2=4, cos⁡(xy)=0x^2=4,\ \cos(xy)=0, so hy=0h_y=0.

hz=2zxh_z=2zx (only z2xz^2x depends on zz). At the point: 2(1)(2)=42(1)(2)=4.

Part (iv): G(x,y)=ex+3ylog⁡(x2+y2)G(x,y)=e^{x+3y}\log(x^2+y^2) at (−1,1)(-1,1).

Gx=ex+3ylog⁡(x2+y2)+ex+3y⋅2xx2+y2=ex+3y[log⁡(x2+y2)+2xx2+y2]G_x = e^{x+3y}\log(x^2+y^2) + e^{x+3y}\cdot\dfrac{2x}{x^2+y^2} = e^{x+3y}\left[\log(x^2+y^2)+\dfrac{2x}{x^2+y^2}\right] (product rule).

At (−1,1)(-1,1): x+3y=−1+3=2x+3y=-1+3=2; x2+y2=1+1=2x^2+y^2=1+1=2; 2xx2+y2=−22=−1\dfrac{2x}{x^2+y^2}=\dfrac{-2}{2}=-1. So Gx=e2[log⁡2−1]G_x=e^2[\log2-1].

Gy=3ex+3ylog⁡(x2+y2)+ex+3y⋅2yx2+y2=ex+3y[3log⁡(x2+y2)+2yx2+y2]G_y = 3e^{x+3y}\log(x^2+y^2) + e^{x+3y}\cdot\dfrac{2y}{x^2+y^2} = e^{x+3y}\left[3\log(x^2+y^2)+\dfrac{2y}{x^2+y^2}\right].

At (−1,1)(-1,1): 2yx2+y2=22=1\dfrac{2y}{x^2+y^2}=\dfrac{2}{2}=1. So Gy=e2[3log⁡2+1]G_y=e^2[3\log2+1].

✓Final answer

(i) fx(2,−5)=27, fy(2,−5)=−14f_x(2,-5)=\boxed{27},\ f_y(2,-5)=\boxed{-14} (ii) gx(1,−2)=11, gy(1,−2)=−4g_x(1,-2)=\boxed{11},\ g_y(1,-2)=\boxed{-4} (iii) hx=2, hy=0, hz=4h_x=\boxed{2},\ h_y=\boxed{0},\ h_z=\boxed{4} (iv) Gx=e2(log⁡2−1)≈−1.29G_x=e^2(\log2-1)\approx\boxed{-1.29}, Gy=e2(3log⁡2+1)≈22.75G_y=e^2(3\log2+1)\approx\boxed{22.75}

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